Selina Concise Solutions for ICSE Class 9 Mathematics Chapter 8 Logarithms

Exercise 8(A)

Question 1. Convert each of the following exponential equations into its logarithmic form:
(i) \( 5^3 = 125 \)
(ii) \( 3^{-2} = \frac{1}{9} \)
(iii) \( 10^{-3} = 0.001 \)
(iv) \( (81)^{\frac{3}{4}} = 27 \)
Answer:
By applying the definition of logarithms, \( a^b = c \)
\( \implies \log_a c = b \):
(i) \( 5^3 = 125 \)
\( \implies \log_5 125 = 3 \)
(ii) \( 3^{-2} = \frac{1}{9} \)
\( \implies \log_3 \frac{1}{9} = -2 \)
(iii) \( 10^{-3} = 0.001 \)
\( \implies \log_{10} 0.001 = -3 \)
(iv) \( (81)^{\frac{3}{4}} = 27 \)
\( \implies \log_{81} 27 = \frac{3}{4} \)
In simple words: Writing an exponent as a logarithm means putting the base of the exponent as the base of the log, while the power becomes the isolated answer.

Exam Tip: Always verify that the base of the exponential form becomes the base of the logarithmic form.

Question 2. Convert each of the following logarithmic equations into its exponential form:
(i) \( \log_8 0.125 = -1 \)
(ii) \( \log_{10} 0.01 = -2 \)
(iii) \( \log_a A = x \)
(iv) \( \log_{10} 1 = 0 \)
Answer:
By converting from logarithmic form using \( \log_a c = b \)
\( \implies a^b = c \):
(i) \( \log_8 0.125 = -1 \)
\( \implies 8^{-1} = 0.125 \)
(ii) \( \log_{10} 0.01 = -2 \)
\( \implies 10^{-2} = 0.01 \)
(iii) \( \log_a A = x \)
\( \implies a^x = A \)
(iv) \( \log_{10} 1 = 0 \)
\( \implies 10^0 = 1 \)
In simple words: To change a logarithm back into an exponent, make the log's base the base of the power, and raise it to the number on the other side.

Exam Tip: Remember that any base raised to the power of 0 is always equal to 1, which explains why \( \log_b 1 = 0 \).

Question 3. Solve for \( x \): \( \log_{10} x = -2 \)
Answer:
We convert the equation into exponential form using \( \log_a c = b \)
\( \implies a^b = c \):
\( \log_{10} x = -2 \)
\( \implies 10^{-2} = x \)
\( \implies x = 10^{-2} \)
\( \implies x = \frac{1}{10^2} \)
\( \implies x = \frac{1}{100} \)
\( \implies x = 0.01 \)
In simple words: Change the logarithm to an exponent to get \( x = 10^{-2} \), then simplify the fraction to find the final decimal value.

Exam Tip: Negative exponents represent reciprocal values, so \( 10^{-2} \) is equal to \( \frac{1}{100} \).

Question 4. Evaluate each of the following:
(i) \( \log_{10} 100 \)
(ii) \( \log_{10} 0.1 \)
(iii) \( \log_{10} 0.001 \)
(iv) \( \log_4 32 \)
(v) \( \log_2 0.125 \)
(vi) \( \log_4 \frac{1}{16} \)
(vii) \( \log_9 27 \)
(viii) \( \log_{27} \frac{1}{81} \)
Answer:
(i) Let \( \log_{10} 100 = x \)
\( \implies 10^x = 100 \)
\( \implies 10^x = 10 \times 10 \)
\( \implies 10^x = 10^2 \)
\( \implies x = 2 \) (since \( a^m = a^n \)
\( \implies m = n \))
Thus, \( \log_{10} 100 = 2 \)

(ii) Let \( \log_{10} 0.1 = x \)
\( \implies 10^x = 0.1 \)
\( \implies 10^x = \frac{1}{10} \)
\( \implies 10^x = 10^{-1} \)
\( \implies x = -1 \) (since \( a^m = a^n \)
\( \implies m = n \))
Thus, \( \log_{10} 0.1 = -1 \)

(iii) Let \( \log_{10} 0.001 = x \)
\( \implies 10^x = 0.001 \)
\( \implies 10^x = \frac{1}{1000} \)
\( \implies 10^x = \frac{1}{10^3} \)
\( \implies 10^x = 10^{-3} \)
\( \implies x = -3 \) (since \( a^m = a^n \)
\( \implies m = n \))
Thus, \( \log_{10} 0.001 = -3 \)

(iv) Let \( \log_4 32 = x \)
\( \implies 4^x = 32 \)
\( \implies (2^2)^x = 2 \times 2 \times 2 \times 2 \times 2 \)
\( \implies 2^{2x} = 2^5 \)
\( \implies 2x = 5 \) (since \( a^m = a^n \)
\( \implies m = n \))
\( \implies x = \frac{5}{2} \)
Thus, \( \log_4 32 = \frac{5}{2} \)

(v) Let \( \log_2 0.125 = x \)
\( \implies 2^x = 0.125 \)
\( \implies 2^x = \frac{125}{1000} \)
\( \implies 2^x = \frac{1}{8} \)
\( \implies 2^x = 8^{-1} \)
\( \implies 2^x = (2 \times 2 \times 2)^{-1} \)
\( \implies 2^x = (2^3)^{-1} \)
\( \implies 2^x = 2^{-3} \)
\( \implies x = -3 \) (since \( a^m = a^n \)
\( \implies m = n \))
Thus, \( \log_2 0.125 = -3 \)

(vi) Let \( \log_4 \frac{1}{16} = x \)
\( \implies 4^x = \frac{1}{16} \)
\( \implies 4^x = \frac{1}{4 \times 4} \)
\( \implies 4^x = (4 \times 4)^{-1} \)
\( \implies 4^x = (4^2)^{-1} \)
\( \implies 4^x = 4^{-2} \)
\( \implies x = -2 \) (since \( a^m = a^n \)
\( \implies m = n \))
Thus, \( \log_4 \frac{1}{16} = -2 \)

(vii) Let \( \log_9 27 = x \)
\( \implies 9^x = 27 \)
\( \implies (3 \times 3)^x = 3 \times 3 \times 3 \)
\( \implies (3^2)^x = (3^3) \)
\( \implies 3^{2x} = (3^3) \)
\( \implies 2x = 3 \) (since \( a^m = a^n \)
\( \implies m = n \))
\( \implies x = \frac{3}{2} \)
Thus, \( \log_9 27 = \frac{3}{2} \)

(viii) Let \( \log_{27} \frac{1}{81} = x \)
\( \implies 27^x = \frac{1}{81} \)
\( \implies (3 \times 3 \times 3)^x = \frac{1}{3 \times 3 \times 3 \times 3} \)
\( \implies (3^3)^x = \frac{1}{3^4} \)
\( \implies (3^3)^x = (3^4)^{-1} \)
\( \implies 3^{3x} = (3^{-4}) \)
\( \implies 3x = -4 \) (since \( a^m = a^n \)
\( \implies m = n \))
\( \implies x = \frac{-4}{3} \)
Thus, \( \log_{27} \frac{1}{81} = \frac{-4}{3} \)
In simple words: Set each logarithm equal to \( x \) and convert it into exponential form. Next, rewrite both sides with a common base and solve for the unknown exponent.

Exam Tip: Expressing decimals as fractions first makes it much easier to identify the common prime base.

Question 5. State whether the following statements are true or false and give reasons:
(i) If \( \log_{10} x = a \), then \( 10^x = a \).
(ii) If \( x^y = z \), then \( \log_z x = y \).
(iii) \( \log_2 8 = 3 \) and \( \log_8 2 = \frac{1}{3} \) are both true.
Answer:
(i) **False**.
Let us analyze the logarithmic relation:
\( \log_{10} x = a \)
\( \implies 10^a = x \)
Hence, the given claim that \( 10^x = a \) is incorrect.

(ii) **False**.
Let us analyze the exponential relation:
\( x^y = z \)
\( \implies \log_x z = y \)
Hence, the given claim that \( \log_z x = y \) is incorrect.

(iii) **True**.
Let us analyze both of the equations:
For the first equation:
\( \log_2 8 = 3 \)
\( \implies 2^3 = 8 \) (which is correct)
For the second equation:
\( \log_8 2 = \frac{1}{3} \)
\( \implies 8^{\frac{1}{3}} = 2 \)
\( \implies (2^3)^{\frac{1}{3}} = 2 \)
\( \implies 2 = 2 \) (which is correct)
Since both statements are verified to be correct, the overall claim is true.
In simple words: Logarithmic forms must translate correctly to exponential forms. Statement (i) and (ii) swap the roles of base, power, and argument incorrectly, while (iii) translates both parts perfectly.

Exam Tip: Always double check your conversion using simple numbers to ensure you have not swapped the base and exponent.

Question 6. Find the value of \( x \) in each of the following:
(i) \( \log_3 x = 0 \)
(ii) \( \log_x 2 = -1 \)
(iii) \( \log_9 243 = x \)
(iv) \( \log_5 (x - 7) = 1 \)
(v) \( \log_4 32 = x - 4 \)
(vi) \( \log_7 (2x^2 - 1) = 2 \)
Answer:
(i) Let us convert the equation:
\( \log_3 x = 0 \)
\( \implies 3^0 = x \)
\( \implies x = 1 \) or \( x = 1 \)

(ii) Let us convert the equation:
\( \log_x 2 = -1 \)
\( \implies x^{-1} = 2 \)
\( \implies \frac{1}{x} = 2 \)
\( \implies x = \frac{1}{2} \)

(iii) Let us convert the equation:
\( \log_9 243 = x \)
\( \implies 9^x = 243 \)
\( \implies (3^2)^x = 3^5 \)
\( \implies 3^{2x} = 3^5 \)
\( \implies 2x = 5 \)
\( \implies x = \frac{5}{2} \)
\( \implies x = 2\frac{1}{2} \)

(iv) Let us convert the equation:
\( \log_5 (x - 7) = 1 \)
\( \implies 5^1 = x - 7 \)
\( \implies 5 = x - 7 \)
\( \implies x = 5 + 7 \)
\( \implies x = 12 \)

(v) Let us convert the equation:
\( \log_4 32 = x - 4 \)
\( \implies 4^{x - 4} = 32 \)
\( \implies (2^2)^{x - 4} = 2^5 \)
\( \implies 2^{2(x - 4)} = 2^5 \)
\( \implies 2x - 8 = 5 \)
\( \implies 2x = 5 + 8 \)
\( \implies 2x = 13 \)
\( \implies x = \frac{13}{2} \)
\( \implies x = 6\frac{1}{2} \)

(vi) Let us convert the equation:
\( \log_7 (2x^2 - 1) = 2 \)
\( \implies 7^2 = 2x^2 - 1 \)
\( \implies 7 \times 7 = 2x^2 - 1 \)
\( \implies 2x^2 - 1 - 49 = 0 \)
\( \implies 2x^2 - 50 = 0 \)
\( \implies 2x^2 = 50 \)
\( \implies x^2 = \frac{50}{2} \)
\( \implies x^2 = 25 \)
\( \implies x = \pm\sqrt{25} \)
\( \implies x = 5 \) (ignoring the negative root)
In simple words: Convert the logarithmic statement into index form. This gives an algebraic equation which you can solve for the unknown variable \( x \).

Exam Tip: When taking the square root, you get positive and negative options, but always verify if the negative option is valid under logarithmic rules.

Question 7. Evaluate each of the following:
(i) \( \log_{10} 0.01 \)
(ii) \( \log_2 \frac{1}{8} \)
(iii) \( \log_5 1 \)
(iv) \( \log_5 125 \)
(v) \( \log_{16} 8 \)
(vi) \( \log_{0.5} 16 \)
Answer:
(i) Let \( \log_{10} 0.01 = x \)
\( \implies 10^x = 0.01 \)
\( \implies 10^x = \frac{1}{100} \)
\( \implies 10^x = \frac{1}{10 \times 10} \)
\( \implies 10^x = \frac{1}{10^2} \)
\( \implies 10^x = 10^{-2} \)
\( \implies x = -2 \)
Hence, \( \log_{10} 0.01 = -2 \)

(ii) Let \( \log_2 \frac{1}{8} = x \)
\( \implies 2^x = \frac{1}{8} \)
\( \implies 2^x = \frac{1}{2 \times 2 \times 2} \)
\( \implies 2^x = \frac{1}{2^3} \)
\( \implies 2^x = 2^{-3} \)
\( \implies x = -3 \)
Hence, \( \log_2 \frac{1}{8} = -3 \)

(iii) Let \( \log_5 1 = x \)
\( \implies 5^x = 1 \)
\( \implies 5^x = 5^0 \)
\( \implies x = 0 \)
Hence, \( \log_5 1 = 0 \)

(iv) Let \( \log_5 125 = x \)
\( \implies 5^x = 125 \)
\( \implies 5^x = 5 \times 5 \times 5 \)
\( \implies 5^x = 5^3 \)
\( \implies x = 3 \)
Hence, \( \log_5 125 = 3 \)

(v) Let \( \log_{16} 8 = x \)
\( \implies 16^x = 8 \)
\( \implies (2 \times 2 \times 2 \times 2)^x = 2 \times 2 \times 2 \)
\( \implies (2^4)^x = 2^3 \)
\( \implies 2^{4x} = 2^3 \)
\( \implies 4x = 3 \)
\( \implies x = \frac{3}{4} \)
Hence, \( \log_{16} 8 = \frac{3}{4} \)

(vi) Let \( \log_{0.5} 16 = x \)
\( \implies 0.5^x = 16 \)
\( \implies \left(\frac{5}{10}\right)^x = 2 \times 2 \times 2 \times 2 \)
\( \implies \left(\frac{1}{2}\right)^x = 2^4 \)
\( \implies \frac{1}{2^x} = 2^4 \)
\( \implies 2^{-x} = 2^4 \)
\( \implies -x = 4 \)
\( \implies x = -4 \)
Hence, \( \log_{0.5} 16 = -4 \)
In simple words: To find the log value, set the entire expression equal to \( x \) and rewrite it exponentially. Next, express both sides with the same base to solve for \( x \).

Exam Tip: When dealing with fractional or decimal bases, express them as powers of integer bases to simplify exponent comparison.

Question 8. If \( \log_a m = n \), express \( a^{n-1} \) in terms of \( a \) and \( m \).
Answer:
Given logarithmic equation:
\( \log_a m = n \)
\( \implies a^n = m \)
Let us divide both sides of this equation by \( a \):
\( \implies \frac{a^n}{a} = \frac{m}{a} \)
\( \implies a^{n-1} = \frac{m}{a} \)
In simple words: Convert the logarithm to exponential form to find \( a^n = m \). Since we want \( a^{n-1} \), divide both sides by \( a \) to lower the exponent by one.

Exam Tip: Remember the basic index rule: dividing powers with identical bases subtracts their exponents, so \( a^n / a = a^{n-1} \).

Question 9. Given that \( \log_2 x = m \) and \( \log_5 y = n \), express the following in terms of \( x \) and \( y \):
(i) \( 2^{m-3} \)
(ii) \( 5^{3n+2} \)
Answer:
From the given equations, we can write:
\( \log_2 x = m \implies 2^m = x \) and \( \log_5 y = n \implies 5^n = y \)

(i) Let us evaluate \( 2^{m-3} \):
\( \implies \frac{2^m}{2^3} = \frac{x}{2^3} \)
\( \implies 2^{m-3} = \frac{x}{8} \)

(ii) Let us evaluate \( 5^{3n+2} \):
\( \implies (5^n)^3 = y^3 \)
\( \implies 5^{3n} = y^3 \)
\( \implies 5^{3n} \times 5^2 = y^3 \times 5^2 \)
\( \implies 5^{3n+2} = 25y^3 \)
In simple words: Change the logarithmic terms to exponents first. For the first part, divide both sides by \( 2^3 \). For the second part, cube both sides first, then multiply by \( 5^2 \).

Exam Tip: Pay attention to exponential operations: cubing adds multiplication to the index, whereas multiplying by a base adds to the index.

Question 10. If \( \log_2 x = a \) and \( \log_3 y = a \), express \( 72^a \) in terms of \( x \) and \( y \).
Answer:
Using the given definitions:
\( \log_2 x = a \implies 2^a = x \) and \( \log_3 y = a \implies 3^a = y \)
The prime factorization of 72 can be written as:
\( 72 = 2 \times 2 \times 2 \times 3 \times 3 \)
Hence,
\( (72)^a = (2 \times 2 \times 2 \times 3 \times 3)^a \)
\( \implies (72)^a = (2^3 \times 3^2)^a \)
\( \implies (72)^a = 2^{3a} \times 3^{2a} \)
\( \implies (72)^a = (2^a)^3 \times (3^a)^2 \)
Substituting the values \( 2^a = x \) and \( 3^a = y \):
\( \implies (72)^a = x^3 y^2 \)
In simple words: Break 72 down into its prime components, 2 and 3. Then raise each part to the power of \( a \) and swap them with \( x \) and \( y \).

Exam Tip: Prime factorization of the base is a powerful first step in simplifying complex logarithmic or exponential relationships.

Question 11. Solve the following logarithmic equation for \( x \): \( \log(x - 1) + \log(x + 1) = \log_2 1 \)
Answer:
We are given the equation:
\( \log(x - 1) + \log(x + 1) = \log_2 1 \)
Since the logarithm of 1 to any base is 0, we can write:
\( \implies \log(x - 1) + \log(x + 1) = 0 \)
Using the product rule of logarithms:
\( \implies \log[(x - 1)(x + 1)] = 0 \)
Converting the equation into exponential form:
\( \implies (x - 1)(x + 1) = 1 \)
\( \implies x^2 - 1 = 1 \)
\( \implies x^2 = 2 \)
\( \implies x = \pm\sqrt{2} \)
As the logarithm of a negative value is not defined, the negative root \( -\sqrt{2} \) must be rejected. Therefore, the only valid solution is \( x = \sqrt{2} \).
In simple words: We can simplify the logarithmic equation by setting the right side to zero because the log of 1 to any base is always zero. Since we cannot take the log of a negative number, we only keep the positive result.

Exam Tip: Always remember to check your final solutions in the original equation to ensure they do not result in taking the logarithm of a non-positive number.

Question 12. Solve the following logarithmic equation for \( x \): \( \log(x^2 - 21) = 2 \)
Answer:
We are given the equation:
\( \log(x^2 - 21) = 2 \)
Converting the equation into exponential form (with base 10):
\( \implies x^2 - 21 = 10^2 \)
\( \implies x^2 - 21 = 100 \)
Adding 21 to both sides:
\( \implies x^2 = 121 \)
Taking the square root of both sides:
\( \implies x = \pm 11 \)
In simple words: Convert the log equation to its exponential form with base 10, then solve for \( x^2 \) to find the square root. Both positive and negative values work here.

Exam Tip: When a log has no specified base, it is usually a common logarithm with base 10. Don't forget that squaring either positive or negative 11 results in 121, so both values are valid.

Exercise 8(B)

Question 1. Express each of the following in terms of \( \log 2 \) and \( \log 3 \), or simplify:
(i) \( \log 36 \)
(ii) \( \log 144 \)
(iii) \( \log 4.5 \)
(iv) \( \log\frac{26}{51} - \log\frac{91}{119} \)
(v) \( \log\frac{75}{16} - 2\log\frac{5}{9} + \log\frac{32}{243} \)
Answer:
(i) Write 36 as prime factors:
\( \log 36 = \log(2 \times 2 \times 3 \times 3) \)
\( = \log(2^2 \times 3^2) \)
Applying the product rule \( \log(mn) = \log m + \log n \):
\( = \log(2^2) + \log(3^2) \)
Using the power rule \( \log(m^n) = n \log m \):
\( = 2\log 2 + 2\log 3 \)

(ii) Find the prime factors of 144:
\( \log 144 = \log(2 \times 2 \times 2 \times 2 \times 3 \times 3) \)
\( = \log(2^4 \times 3^2) \)
Using the product rule:
\( = \log(2^4) + \log(3^2) \)
Using the power rule:
\( = 4\log 2 + 2\log 3 \)

(iii) First, write 4.5 as a fraction:
\( \log 4.5 = \log \frac{45}{10} \)
Reduce the fraction by dividing the numerator and denominator by 5:
\( = \log \frac{5 \times 3 \times 3}{5 \times 2} \)
\( = \log \frac{3^2}{2} \)
Apply the quotient rule \( \log\frac{m}{n} = \log m - \log n \):
\( = \log(3^2) - \log 2 \)
Apply the power rule:
\( = 2\log 3 - \log 2 \)

(iv) We have:
\( \log\frac{26}{51} - \log\frac{91}{119} \)
Apply the quotient rule:
\( = \log \left( \frac{26/51}{91/119} \right) \)
Simplify the fraction division:
\( = \log \left( \frac{26}{51} \times \frac{119}{91} \right) \)
Factoring the terms to simplify:
\( = \log \left( \frac{2 \times 13}{3 \times 17} \times \frac{7 \times 17}{7 \times 13} \right) \)
Cancelling out common factors (13, 17, and 7):
\( = \log \frac{2}{3} \)
Applying the division rule:
\( = \log 2 - \log 3 \)

(v) We have:
\( \log\frac{75}{16} - 2\log\frac{5}{9} + \log\frac{32}{243} \)
First, apply the power rule to the middle term:
\( = \log\frac{75}{16} - \log\left(\frac{5}{9}\right)^2 + \log\frac{32}{243} \)
\( = \log\frac{75}{16} - \log\left(\frac{5 \times 5}{9 \times 9}\right) + \log\frac{32}{243} \)
\( = \log\frac{75}{16} - \log\frac{25}{81} + \log\frac{32}{243} \)
Combine the first two terms using the quotient rule:
\( = \log\left( \frac{75/16}{25/81} \right) + \log\frac{32}{243} \)
\( = \log\left( \frac{75}{16} \times \frac{81}{25} \right) + \log\frac{32}{243} \)
Simplify the product:
\( = \log\left( \frac{3 \times 25}{16} \times \frac{81}{25} \right) + \log\frac{32}{243} \)
\( = \log\frac{3 \times 81}{16} + \log\frac{32}{243} \)
\( = \log\frac{243}{16} + \log\frac{32}{243} \)
Combine the terms using the product rule:
\( = \log\left( \frac{243}{16} \times \frac{32}{243} \right) \)
Simplify the fraction:
\( = \log\frac{32}{16} \)
\( = \log 2 \)
In simple words: We can simplify complex logarithmic terms by breaking down numbers into their prime factors (like 2 and 3) or by using standard rules for multiplying, dividing, and raising powers in logarithms.

Exam Tip: When simplifying logs with fractions, it is often helpful to convert division into subtraction and multiplication into addition, making it much easier to group similar terms.

Question 2. Express each of the following equations in a form free from logarithms:
(i) \( 2\log x - \log y = 1 \)
(ii) \( 2\log x + 3\log y = \log a \)
(iii) \( a\log x - b\log y = 2\log 3 \)
Answer:
(i) Given:
\( 2\log x - \log y = 1 \)
Apply the power rule to the first term:
\( \implies \log x^2 - \log y = 1 \)
Since \( \log 10 = 1 \), we can write:
\( \implies \log \frac{x^2}{y} = \log 10 \)
Removing the logarithms from both sides:
\( \implies \frac{x^2}{y} = 10 \)
\( \implies x^2 = 10y \)

(ii) Given:
\( 2\log x + 3\log y = \log a \)
Using the power rule:
\( \implies \log x^2 + \log y^3 = \log a \)
Apply the product rule:
\( \implies \log(x^2 y^3) = \log a \)
Removing the logarithms:
\( \implies x^2 y^3 = a \)

(iii) Given:
\( a\log x - b\log y = 2\log 3 \)
Apply the power rule:
\( \implies \log x^a - \log y^b = \log 3^2 \)
Apply the quotient rule:
\( \implies \log \frac{x^a}{y^b} = \log 9 \)
Removing the logarithms:
\( \implies \frac{x^a}{y^b} = 9 \)
\( \implies x^a = 9y^b \)
In simple words: To remove logarithms from an equation, use log properties to combine all terms on each side into a single log, then equate the expressions inside the logs.

Exam Tip: Remember that a constant like 1 can be replaced with \( \log_{10} 10 \) to make it easy to combine with other log terms.

Question 3. Evaluate the following logarithmic expressions:
(i) \( \log 5 + \log 8 - 2\log 2 \)
(ii) \( \log_{10} 8 + \log_{10} 25 + 2\log_{10} 3 - \log_{10} 18 \)
(iii) \( \log 4 + \frac{1}{3}\log 125 - \frac{1}{5}\log 32 \)
Answer:
(i) Given expression:
\( \log 5 + \log 8 - 2\log 2 \)
Apply the power rule to the subtraction term:
\( = \log 5 + \log 8 - \log 2^2 \)
Combine the terms with addition using the product rule:
\( = \log(5 \times 8) - \log 4 \)
\( = \log 40 - \log 4 \)
Apply the quotient rule to simplify the subtraction:
\( = \log \frac{40}{4} \)
\( = \log 10 \)
Since \( \log 10 = 1 \):
\( = 1 \)

(ii) Given expression:
\( \log_{10} 8 + \log_{10} 25 + 2\log_{10} 3 - \log_{10} 18 \)
Using the power rule on the third term:
\( = \log_{10} 8 + \log_{10} 25 + \log_{10} 3^2 - \log_{10} 18 \)
\( = \log_{10} 8 + \log_{10} 25 + \log_{10} 9 - \log_{10} 18 \)
Combine the positive log terms using the product rule:
\( = \log_{10} (8 \times 25 \times 9) - \log_{10} 18 \)
\( = \log_{10} 1800 - \log_{10} 18 \)
Apply the quotient rule:
\( = \log_{10} \frac{1800}{18} \)
\( = \log_{10} 100 \)
Since \( 100 = 10^2 \):
\( = 2 \)

(iii) Given expression:
\( \log 4 + \frac{1}{3}\log 125 - \frac{1}{5}\log 32 \)
Express the terms using fractional powers:
\( = \log 4 + \log (125)^{\frac{1}{3}} - \log (32)^{\frac{1}{5}} \)
Simplify the roots of the numbers:
\( = \log 4 + \log (5^3)^{\frac{1}{3}} - \log (2^5)^{\frac{1}{5}} \)
\( = \log 4 + \log 5 - \log 2 \)
Combine using the product rule:
\( = \log(4 \times 5) - \log 2 \)
\( = \log 20 - \log 2 \)
Using the quotient rule:
\( = \log \frac{20}{2} \)
\( = \log 10 \)
\( = 1 \)
In simple words: We can simplify these expressions by converting coefficients into exponents (like square roots or cube roots) and then combining the terms using standard log laws to get a single number.

Exam Tip: Fractional coefficients like \( \frac{1}{3} \) and \( \frac{1}{5} \) represent cube roots and fifth roots, respectively. Recognizing perfect powers (like \( 125 = 5^3 \)) makes these very easy to simplify.

Question 4. Prove that: \( 2\log\frac{15}{18} - \log\frac{25}{162} + \log\frac{4}{9} = \log 2 \)
Answer:
We will start by simplifying the Left Hand Side (L.H.S.):
\( L.H.S. = 2\log\frac{15}{18} - \log\frac{25}{162} + \log\frac{4}{9} \)
Apply the power rule to the first term:
\( = \log\left(\frac{15}{18}\right)^2 - \log\frac{25}{162} + \log\frac{4}{9} \)
\( = \log\left[\left(\frac{15}{18}\right) \times \left(\frac{15}{18}\right)\right] - \log\frac{25}{162} + \log\frac{4}{9} \)
Combine the terms being added using the product rule:
\( = \log\left[\left(\frac{15}{18}\right) \times \left(\frac{15}{18}\right) \times \frac{4}{9}\right] - \log\frac{25}{162} \)
Now apply the quotient rule to the subtraction term:
\( = \log \left( \frac{\left(\frac{15}{18}\right) \times \left(\frac{15}{18}\right) \times \frac{4}{9}}{\frac{25}{162}} \right) \)
Convert the division into multiplication by the reciprocal of the divisor:
\( = \log\left[ \left(\frac{15}{18}\right) \times \left(\frac{15}{18}\right) \times \frac{4}{9} \times \frac{162}{25} \right] \)
Let's simplify the product of the fractions:
First, \( \frac{15}{18} = \frac{5}{6} \).
So we have:
\( = \log\left[ \frac{5}{6} \times \frac{5}{6} \times \frac{4}{9} \times \frac{162}{25} \right] \)
\( = \log\left[ \frac{25}{36} \times \frac{4 \times 18}{25} \right] \) (since \( 162 = 9 \times 18 \))
\( = \log\left[ \frac{25}{36} \times \frac{72}{25} \right] \)
Cancel the common factors of 25 and divide 72 by 36:
\( = \log \frac{72}{36} \)
\( = \log 2 \)
This equals the Right Hand Side (R.H.S.).
Hence proved.
In simple words: We start with the left side and use the power rule to move the coefficient 2 to the exponent. Then we merge the logs using the addition and subtraction laws, simplify the resulting fraction, and show it is equal to \( \log 2 \).

Exam Tip: Be careful when multiplying and dividing large fractions. Simplifying individual fractions first, like turning \( \frac{15}{18} \) into \( \frac{5}{6} \), can make the arithmetic much cleaner.

Question 5. Solve the following equation for \( x \): \( x - \log 48 + 3\log 2 = \frac{1}{3}\log 125 - \log 3 \)
Answer:
We are given:
\( x - \log 48 + 3\log 2 = \frac{1}{3}\log 125 - \log 3 \)
Isolate \( x \) on the Left Hand Side (L.H.S.):
\( \implies x = \frac{1}{3}\log 125 - \log 3 + \log 48 - 3\log 2 \)
Apply the power rule to simplify the coefficients:
\( \implies x = \log(125)^{\frac{1}{3}} - \log 3 + \log 48 - \log 2^3 \)
Find the cube root of 125 and simplify \( 2^3 \):
\( \implies x = \log(5^3)^{\frac{1}{3}} - \log 3 + \log 48 - \log 8 \)
\( \implies x = \log 5 - \log 3 + \log 48 - \log 8 \)
Rearrange the terms to group positive and negative terms:
\( \implies x = \log 5 + \log 48 - \log 3 - \log 8 \)
Factor out the negative sign:
\( \implies x = (\log 5 + \log 48) - (\log 3 + \log 8) \)
Combine terms inside parentheses using the product rule:
\( \implies x = \log(5 \times 48) - \log(3 \times 8) \)
Now use the quotient rule:
\( \implies x = \log \frac{5 \times 48}{3 \times 8} \)
Simplify the fraction inside the log:
\( \implies x = \log \frac{240}{24} \)
\( \implies x = \log 10 \)
Since \( \log 10 = 1 \):
\( \implies x = 1 \)
In simple words: Move all logarithmic terms to one side of the equation to isolate \( x \). Then, combine them into a single log term using the rules of logs, which simplifies nicely to \( \log 10 \), giving \( x = 1 \).

Exam Tip: When grouping terms, putting positive terms and negative terms together makes it easier to apply the product and quotient rules correctly.

Question 6. Express the following as a single logarithm: \( \log_{10} 2 + 1 \)
Answer:
Given expression:
\( \log_{10} 2 + 1 \)
Since \( 1 = \log_{10} 10 \), substitute this into the expression:
\( = \log_{10} 2 + \log_{10} 10 \)
Applying the product rule of logarithms:
\( = \log_{10} (2 \times 10) \)
\( = \log_{10} 20 \)
In simple words: We can turn the number 1 into \( \log_{10} 10 \) because 10 raised to the power of 1 is 10. Then we combine the two logs by multiplying their values.

Exam Tip: To add a constant to a log, write that constant as a logarithm with the same base so they can be merged.

Question 7. Solve the following equations for \( x \):
(i) \( \log_{10}(x - 10) = 1 \)
(ii) \( \log(x^2 - 21) = 2 \)
(iii) \( \log(x - 2) + \log(x + 2) = \log 5 \)
(iv) \( \log(x + 5) + \log(x - 5) = 4\log 2 + 2\log 3 \)
Answer:
(i) Given:
\( \log_{10}(x - 10) = 1 \)
Since \( \log_{10} 10 = 1 \):
\( \implies \log_{10} (x - 10) = \log_{10} 10 \)
Comparing the arguments on both sides:
\( \implies x - 10 = 10 \)
\( \implies x = 10 + 10 \)
\( \implies x = 20 \)

(ii) Given:
\( \log(x^2 - 21) = 2 \)
Since the base is 10, write 2 as \( \log 100 \):
\( \implies \log(x^2 - 21) = \log 100 \)
Comparing the terms inside the logs:
\( \implies x^2 - 21 = 100 \)
Subtract 100 from both sides:
\( \implies x^2 - 21 - 100 = 0 \)
\( \implies x^2 - 121 = 0 \)
Rearranging:
\( \implies x^2 = 121 \)
Taking the square root of both sides:
\( \implies x = \pm\sqrt{121} \)
\( \implies x = \pm 11 \)

(iii) Given:
\( \log(x - 2) + \log(x + 2) = \log 5 \)
Applying the product rule to the Left Hand Side:
\( \implies \log\{(x - 2)(x + 2)\} = \log 5 \)
Expand the product of conjugates:
\( \implies \log(x^2 - 4) = \log 5 \)
Comparing the arguments of both logs:
\( \implies x^2 - 4 = 5 \)
\( \implies x^2 = 9 \)
Taking square roots:
\( \implies x = \pm\sqrt{9} \)
\( \implies x = \pm\sqrt{3^2} \)
\( \implies x = \pm 3 \)

(iv) Given:
\( \log(x + 5) + \log(x - 5) = 4\log 2 + 2\log 3 \)
Combine the terms on the Left Hand Side using the product rule:
\( \implies \log\{(x + 5)(x - 5)\} = 4\log 2 + 2\log 3 \)
Using the power rule on the Right Hand Side:
\( \implies \log(x^2 - 25) = \log 2^4 + \log 3^2 \)
\( \implies \log(x^2 - 25) = \log 16 + \log 9 \)
Apply the product rule on the Right Hand Side:
\( \implies \log(x^2 - 25) = \log (16 \times 9) \)
\( \implies \log(x^2 - 25) = \log 144 \)
Equating the arguments:
\( \implies x^2 - 25 = 144 \)
\( \implies x^2 = 144 + 25 \)
\( \implies x^2 = 169 \)
\( \implies x = \pm\sqrt{169} \)
\( \implies x = \pm\sqrt{13^2} \)
\( \implies x = \pm 13 \)
In simple words: We can solve logarithmic equations by first combining the log terms on each side into a single log. Once we have a single log on both sides, we can equate what is inside them and solve the remaining algebraic equation for \( x \).

Exam Tip: Pay attention to whether your textbook expects you to reject negative values of \( x \) that might result in negative arguments for the original logarithms. In some boards/exercises, all mathematical roots are presented, while in others, only values satisfying the domain of the original logs are accepted.

Question 8. Solve the following equations for \( x \):
(i) \( \frac{\log 81}{\log 27} = x \)
(ii) \( \frac{\log 128}{\log 32} = x \)
(iii) \( \frac{\log 64}{\log 8} = \log x \)
(iv) \( \frac{\log 225}{\log 15} = \log x \)
Answer:
(i) Given:
\( \frac{\log 81}{\log 27} = x \)
Isolating \( x \) on the Left Hand Side:
\( \implies x = \frac{\log 81}{\log 27} \)
Expressing the numbers in terms of base 3:
\( \implies x = \frac{\log(3 \times 3 \times 3 \times 3)}{\log(3 \times 3 \times 3)} \)
\( \implies x = \frac{\log 3^4}{\log 3^3} \)
Using the power rule:
\( \implies x = \frac{4\log 3}{3\log 3} \)
Cancelling the common factor of \( \log 3 \):
\( \implies x = \frac{4}{3} \)
\( \implies x = 1\frac{1}{3} \)

(ii) Given:
\( \frac{\log 128}{\log 32} = x \)
Isolating \( x \):
\( \implies x = \frac{\log 128}{\log 32} \)
Expressing the numbers as powers of 2:
\( \implies x = \frac{\log(2 \times 2 \times 2 \times 2 \times 2 \times 2 \times 2)}{\log(2 \times 2 \times 2 \times 2 \times 2)} \)
\( \implies x = \frac{\log 2^7}{\log 2^5} \)
Using the power rule:
\( \implies x = \frac{7\log 2}{5\log 2} \)
Cancelling the common factor \( \log 2 \):
\( \implies x = \frac{7}{5} \)
\( \implies x = 1.4 \)

(iii) Given:
\( \frac{\log 64}{\log 8} = \log x \)
Rearranging the terms:
\( \implies \log x = \frac{\log 64}{\log 8} \)
Expressing the numbers in terms of base 2:
\( \implies \log x = \frac{\log(2 \times 2 \times 2 \times 2 \times 2 \times 2)}{\log(2 \times 2 \times 2)} \)
\( \implies \log x = \frac{\log 2^6}{\log 2^3} \)
Using the power rule:
\( \implies \log x = \frac{6\log 2}{3\log 2} \)
Cancelling out the common factor \( \log 2 \):
\( \implies \log x = \frac{6}{3} \)
\( \implies \log x = 2 \)
Assuming common logarithm with base 10:
\( \implies \log_{10} x = 2 \)
Writing in exponential form:
\( \implies 10^2 = x \)
\( \implies x = 10 \times 10 \)
\( \implies x = 100 \)

(iv) Given:
\( \frac{\log 225}{\log 15} = \log x \)
Isolating \( \log x \) on the Left Hand Side:
\( \implies \log x = \frac{\log 225}{\log 15} \)
Since \( 225 = 15 \times 15 \):
\( \implies \log x = \frac{\log (15 \times 15)}{\log 15} \)
\( \implies \log x = \frac{\log 15^2}{\log 15} \)
Using the power rule:
\( \implies \log x = \frac{2\log 15}{\log 15} \)
Cancelling out the common factor \( \log 15 \):
\( \implies \log x = 2 \)
Assuming common logarithm base 10:
\( \implies \log_{10} x = 2 \)
Converting to exponential form:
\( \implies 10^2 = x \)
\( \implies x = 10 \times 10 \)
\( \implies x = 100 \)
In simple words: For these equations, write the numbers inside the logarithms as powers of a common base. This allows you to pull the exponents out, cancel the common log factors, and simplify the fraction to find the value of \( x \).

Exam Tip: When you have a fraction with logs, such as \( \frac{\log A}{\log B} \), look for a common base for \( A \) and \( B \) (like 3 for 81 and 27) so that you can cancel out the log term.

Question 9. Given that \(\log x = m + n\) and \(\log y = m - n\), express \(\log \frac{10x}{y^2}\) in terms of \(m\) and \(n\).
Answer:
We are given:
\(\log x = m + n\)
\(\log y = m - n\)

Let us analyze the expression \(\log \frac{10x}{y^2}\):
\(\log \frac{10x}{y^2} = \log 10x - \log y^2\)
\(= \log 10x - 2\log y\) [Since \(\log_a m^n = n\log_a m\)]
\(= \log 10 + \log x - 2\log y\) [Since \(\log_a m + \log_a n = \log_a (mn)\)]
\(= 1 + \log x - 2\log y\)
\(= 1 + (m + n) - 2(m - n)\)
\(= 1 + m + n - 2m + 2n\)

\(\implies \log \frac{10x}{y^2} = 1 - m + 3n\)
In simple words: First, break down the fraction using logarithmic subtraction and addition rules. Then, substitute the values of \(\log x\) and \(\log y\) with the given expressions in terms of \(m\) and \(n\), and simplify the final equation.

Exam Tip: Remember to apply the quotient rule first, then the power and product rules. Be careful with signs when distributing the negative coefficient across the brackets during substitution.

Question 10. State whether the following statements are true or false. Justify your answer.
(i) \(\log 1 \times \log 1000 = 0\)
(ii) \(\frac{\log x}{\log y} = \log x - \log y\)
(iii) If \(\frac{\log 25}{\log 5} = \log x\), then \(x = 2\).
(iv) \(\log x + \log y = \log x \times \log y\)
Answer:
(i) We observe that:
\(\log 1 = 0\) and \(\log 1000 = 3\)
\(\therefore \log 1 \times \log 1000 = 0 \times 3 = 0\)
Hence, the statement \(\log 1 \times \log 1000 = 0\) is true.

(ii) By the quotient property of logarithms:
\(\log\left(\frac{m}{n}\right) = \log m - \log n\)
\(\therefore \frac{\log x}{\log y} \neq \log x - \log y\)
Hence, the statement \(\frac{\log x}{\log y} = \log x - \log y\) is false.

(iii) We are given:
\(\frac{\log 25}{\log 5} = \log x\)

\(\implies \frac{\log (5 \times 5)}{\log 5} = \log x\)

\(\implies \frac{\log 5^2}{\log 5} = \log x\)

\(\implies \frac{2\log 5}{\log 5} = \log x\) [Since \(\log_a m^n = n\log_a m\)]

\(\implies 2 = \log_{10} x\)

\(\implies 10^2 = x\)

\(\implies x = 100\)
Hence, the statement \(x = 2\) is false.

(iv) According to log rules:
\(\log x + \log y = \log (xy)\)
\(\therefore \log x + \log y \neq \log x \times \log y\)
Hence, the statement \(\log x + \log y = \log x \times \log y\) is false.
In simple words: Logarithmic operations have specific properties: addition corresponds to multiplying terms inside a single log, subtraction corresponds to dividing them, and division of two separate logs cannot be simplified into subtraction.

Exam Tip: Never confuse the quotient rule \(\log \frac{x}{y} = \log x - \log y\) with the division of two logs \(\frac{\log x}{\log y}\), which does not simplify in the same way.

Question 11. Given that \(\log_{10} 2 = a\) and \(\log_{10} 3 = b\), find the values of the following in terms of \(a\) and \(b\):
(i) \(\log 12\)
(ii) \(\log 2.25\)
(iii) \(\log 2\frac{1}{4}\)
(iv) \(\log 5.4\)
(v) \(\log 60\)
(vi) \(\log 3\frac{1}{8}\)
Answer:
(i) We can write \(\log 12\) as:
\(\log 12 = \log (2 \times 2 \times 3)\)
\(= \log (2 \times 2) + \log 3\) [Since \(\log_a (mn) = \log_a m + \log_a n\)]
\(= \log 2^2 + \log 3\)
\(= 2\log 2 + \log 3\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2a + b\) [Given that \(\log_{10} 2 = a\) and \(\log_{10} 3 = b\)]

(ii) Converting the decimal to a fraction:
\(\log 2.25 = \log \frac{225}{100}\)
\(= \log \frac{25 \times 9}{25 \times 4}\)
\(= \log \frac{9}{4}\)
\(= \log \left(\frac{3}{2}\right)^2\)
\(= 2\log \left(\frac{3}{2}\right)\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2(\log 3 - \log 2)\) [Since \(\log_a \frac{m}{n} = \log_a m - \log_a n\)]
\(= 2(b - a)\) [Given that \(\log_{10} 2 = a\) and \(\log_{10} 3 = b\)]
\(= 2b - 2a\)

(iii) Expressing the mixed fraction as an improper fraction:
\(\log 2\frac{1}{4} = \log \frac{9}{4}\)
\(= \log \left(\frac{3}{2}\right)^2\)
\(= 2\log \left(\frac{3}{2}\right)\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2(\log 3 - \log 2)\) [Since \(\log_a \frac{m}{n} = \log_a m - \log_a n\)]
\(= 2(b - a)\) [Given that \(\log_{10} 2 = a\) and \(\log_{10} 3 = b\)]
\(= 2b - 2a\)

(iv) Converting the decimal to a fraction:
\(\log 5.4 = \log \frac{54}{10}\)
\(= \log \left(\frac{2 \times 3 \times 3 \times 3}{10}\right)\)
\(= \log (2 \times 3 \times 3 \times 3) - \log_{10} 10\) [Since \(\log_a \frac{m}{n} = \log_a m - \log_a n\)]
\(= \log_{10} 2 + \log_{10} 3^3 - \log_{10} 10\) [Since \(\log_a (mn) = \log_a m + \log_a n\)]
\(= \log_{10} 2 + 3\log_{10} 3 - \log_{10} 10\) [Since \(\log_a m^n = n\log_a m\)]
\(= \log_{10} 2 + 3\log_{10} 3 - 1\) [Since \(\log_{10} 10 = 1\)]
\(= a + 3b - 1\) [Given that \(\log_{10} 2 = a\) and \(\log_{10} 3 = b\)]

(v) Factoring the number:
\(\log 60 = \log_{10} (10 \times 2 \times 3)\)
\(= \log_{10} 10 + \log_{10} 2 + \log_{10} 3\) [Since \(\log_a (mn) = \log_a m + \log_a n\)]
\(= 1 + \log_{10} 2 + \log_{10} 3\) [Since \(\log_{10} 10 = 1\)]
\(= 1 + a + b\) [Given that \(\log_{10} 2 = a\) and \(\log_{10} 3 = b\)]

(vi) Converting the mixed fraction and simplifying:
\(\log 3\frac{1}{8} = \log_{10} \left(\frac{25}{8} \times \frac{4}{4}\right)\)
\(= \log_{10} \left(\frac{100}{32}\right)\)
\(= \log_{10} 100 - \log_{10} 32\) [Since \(\log_a \frac{m}{n} = \log_a m - \log_a n\)]
\(= \log_{10} 100 - \log_{10} 2^5\)
\(= 2 - \log_{10} 2^5\) [Since \(\log_{10} 100 = 2\)]
\(= 2 - 5\log_{10} 2\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2 - 5a\) [Since \(\log_{10} 2 = a\)]
In simple words: Break down each number into prime factors like 2 and 3, or use powers of 10. Then substitute the given variables \(a\) and \(b\) to represent the final simplified values.

Exam Tip: Try to express decimal fractions in terms of prime factors or multiples of 10, as bases of 10 evaluate cleanly to integers like 1 or 2.

Question 12. Given that \(\log 2 = 0.3010\) and \(\log 3 = 0.4771\), find the values of:
(i) \(\log 12\)
(ii) \(\log 1.2\)
(iii) \(\log 3.6\)
(iv) \(\log 15\)
(v) \(\log 25\)
(vi) \(\frac{2}{3} \log 8\)
Answer:
(i) Factoring the argument:
\(\log 12 = \log (2 \times 2 \times 3)\)
\(= \log 2 \times 2 + \log 3\) [Since \(\log_a (mn) = \log_a m + \log_a n\)]
\(= \log 2^2 + \log 3\)
\(= 2\log 2 + \log 3\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2(0.3010) + 0.4771\) [Given \(\log 2 = 0.3010\) and \(\log 3 = 0.4771\)]
\(= 1.0791\)

(ii) Rewriting as a fraction:
\(\log 1.2 = \log \frac{12}{10}\)
\(= \log 12 - \log 10\) [Since \(\log_a \frac{m}{n} = \log_a m - \log_a n\)]
\(= \log (2 \times 2 \times 3) - 1\) [Since \(\log 10 = 1\)]
\(= \log (2 \times 2) + \log 3 - 1\) [Since \(\log_a (mn) = \log_a m + \log_a n\)]
\(= \log 2^2 + \log 3 - 1\)
\(= 2\log 2 + \log 3 - 1\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2(0.3010) + 0.4771 - 1\) [Given \(\log 2 = 0.3010\) and \(\log 3 = 0.4771\)]
\(= 1.0791 - 1\)
\(= 0.0791\)

(iii) Expressing as a fraction:
\(\log 3.6 = \log \frac{36}{10}\)
\(= \log 36 - \log 10\) [Since \(\log_a \frac{m}{n} = \log_a m - \log_a n\)]
\(= \log (2 \times 2 \times 3 \times 3) - 1\) [Since \(\log 10 = 1\)]
\(= \log (2 \times 2) + \log (3 \times 3) - 1\) [Since \(\log_a (mn) = \log_a m + \log_a n\)]
\(= \log 2^2 + \log 3^2 - 1\)
\(= 2\log 2 + 2\log 3 - 1\) [Since \(\log_a m^n = n\log_a m\)]
\(= 2(0.3010) + 2(0.4771) - 1\) [Given \(\log 2 = 0.3010\) and \(\log 3 = 0.4771\)]
\(= 1.5562 - 1\)
\(= 0.5562\)

(iv) Rewriting with factors of 10 and 2:
\(\log 15 = \log \left(\frac{15}{10} \times 10\right)\)
\(= \log \left(\frac{15}{10}\right) + \log 10\)
\(= \log \left(\frac{3}{2}\right) + 1\) [Since \(\log 10 = 1\)]
\(= \log 3 - \log 2 + 1\) [Since \(\log \frac{m}{n} = \log m - \log n\)]
\(= 0.4771 - 0.3010 + 1\)
\(= 1.1761\)

(v) Expressing in terms of powers of 10:
\(\log 25 = \log \left(\frac{25}{4} \times 4\right)\)
\(= \log \left(\frac{100}{4}\right)\) [Since \(\log (mn) = \log m + \log n\)]
\(= \log 100 - \log (2 \times 2)\) [Since \(\log \frac{m}{n} = \log m - \log n\)]
\(= 2 - \log (2^2)\) [Since \(\log 100 = 2\)]
\(= 2 - 2\log 2\) [Since \(\log m^n = n\log m\)]
\(= 2 - 2(0.3010)\)
\(= 1.398\)

(vi) Simplifying the logarithm:
\(\frac{2}{3}\log 8 = \frac{2}{3}\log (2 \times 2 \times 2)\)
\(= \frac{2}{3}\log 2^3\)
\(= 3 \times \frac{2}{3}\log 2\) [Since \(\log m^n = n\log m\)]
\(= 2\log 2\)
\(= 2 \times 0.3010\)
\(= 0.602\)
In simple words: Convert the numbers inside the logarithms to combinations of 2, 3, and 10. Then substitute the given values to calculate the decimal results.

Exam Tip: Be comfortable with standard conversions like writing \(\log 15\) as \(\log (30/2)\) or \(\log (3/2 \times 10)\) to make calculation with \(\log 2\) and \(\log 3\) straightforward.

Question 13. (i) Find the value of \(x\) that satisfies the equation \(2\log_{10} x + 1 = \log_{10} 250\).
(ii) Using the value of \(x\) from part (i), find the value of \(\log_{10} 2x\).
Answer:
(i) Let us solve the given equation:
\(2\log_{10} x + 1 = \log_{10} 250\)

\(\implies \log_{10} x^2 + 1 = \log_{10} 250\) [Since \(\log_a m^n = n\log_a m\)]

\(\implies \log_{10} x^2 + \log_{10} 10 = \log_{10} 250\) [Since \(\log_{10} 10 = 1\)]

\(\implies \log_{10} (x^2 \times 10) = \log_{10} 250\) [Since \(\log_a m + \log_a n = \log_a (mn)\)]

\(\implies x^2 \times 10 = 250\)

\(\implies x^2 = 25\)

\(\implies x = \sqrt{25}\)

\(\implies x = 5\)

(ii) From the calculation in part (i), we have \(x = 5\).
Substituting this value into the expression:
\(\log_{10} 2x = \log_{10} (2 \times 5)\)
\(= \log_{10} 10\)
\(= 1\) [Since \(\log_{10} 10 = 1\)]
In simple words: First, rewrite the constant 1 as \(\log_{10} 10\) and combine the terms on the left side. Compare the inputs of both logarithms to solve for \(x\), then plug \(x = 5\) into \(\log_{10} 2x\) to get the final value.

Exam Tip: Always convert naked constants like \(1\) or \(2\) into logarithmic terms like \(\log_{10} 10\) or \(\log_{10} 100\) to easily merge them with other log terms.

Question 14. If \(3\log x + \frac{1}{2}\log y = 2\), express \(y\) in terms of \(x\).
Answer:
We are given:
\(3\log x + \frac{1}{2}\log y = 2\)

\(\implies \log x^3 + \log \sqrt{y} = 2\)

\(\implies \log (x^3\sqrt{y}) = 2\)

\(\implies x^3\sqrt{y} = 10^2\)

\(\implies \sqrt{y} = \frac{10^2}{x^3}\)
By taking the square of both sides, we get:
\(y = \frac{10000}{x^6}\)

\(\implies y = 10000x^{-6}\)
In simple words: Combine the logarithmic terms on the left side by moving the coefficients to exponents and multiplying. Then convert the logarithm to an exponential equation and isolate \(y\).

Exam Tip: Remember that squaring a fraction like \(\frac{100}{x^3}\) requires squaring both the numerator and the denominator, leading to \(\frac{10000}{x^6}\).

Question 15. Given that \(x = (100)^a\), \(y = (10000)^b\), and \(z = (10)^c\), express the value of \(\log \frac{10\sqrt{y}}{x^2z^3}\) in terms of \(a\), \(b\), and \(c\).
Answer:
We are given:
\(x = (100)^a\), \(y = (10000)^b\), and \(z = (10)^c\)
Taking logarithms on both sides, we get:
\(\log x = a \log 100\), \(\log y = b \log 10000\), and \(\log z = c \log 10\)

Let us write down the given expression:
\(\log \frac{10\sqrt{y}}{x^2 z^3} = \log (10\sqrt{y}) - \log (x^2 z^3)\)
\(= \log (10y^{1/2}) - \log x^2 - \log z^3\)
\(= \log 10 + \log y^{1/2} - \log x^2 - \log z^3\)
\(= \log 10 + \frac{1}{2}\log y - 2\log x - 3\log z\)
\(= 1 + \frac{1}{2}\log (10000)^b - 2\log (100)^a - 3\log (10)^c\) [Since \(\log 10 = 1\)]
\(= 1 + \frac{b}{2}\log(10)^4 - a\log(10)^2 - 3c\log 10\)
\(= 1 + \frac{b}{2} \times 4\log 10 - 2 \times 2a\log 10 - 3c\log 10\)
\(= 1 + 2b - 4a - 3c\)
In simple words: Expand the logarithmic expression using division, multiplication, and power rules. Replace the variables with their powers-of-10 forms, and simplify using the fact that \(\log_{10} 10 = 1\).

Exam Tip: Be precise when converting bases to 10 (e.g., \(100 = 10^2\) and \(10000 = 10^4\)) to make the final algebraic reduction straightforward.

Question 16. Solve the following equation for \(x\):
\(3(\log 5 - \log 3) - (\log 5 - 2\log 6) = 2 - \log x\).
Answer:
We are given the equation:
\(3(\log 5 - \log 3) - (\log 5 - 2\log 6) = 2 - \log x\)

\(\implies 3\log 5 - 3\log 3 - \log 5 + 2\log(2 \times 3) = 2 - \log x\)

\(\implies 3\log 5 - 3\log 3 - \log 5 + 2\log 2 + 2\log 3 = 2 - \log x\)

\(\implies 2\log 5 - \log 3 + 2\log 2 = 2 - \log x\)

\(\implies 2\log 5 - \log 3 + 2\log 2 + \log x = 2\)

\(\implies \log 5^2 - \log 3 + \log 2^2 + \log x = 2\)

\(\implies \log \left(\frac{25 \times 4 \times x}{3}\right) = 2\)

\(\implies \log \left(\frac{100x}{3}\right) = 2\)

\(\implies \frac{100x}{3} = 10^2\)

\(\implies \frac{x}{3} = 1\)

\(\implies x = 3\)
In simple words: Group and simplify the logarithmic terms on one side by factoring and combining. Once you get a single log term equal to 2, convert it to exponential form to find the value of \(x\).

Exam Tip: When dealing with coefficients in front of parenthesis, expand carefully to ensure the negative sign is applied correctly to every term inside.

Exercise 8(C)

Question 1. Given that \(\log_{10} 8 = 0.90\), find the values of:
(i) \(\log 4\)
(ii) \(\log \sqrt{32}\)
Answer:
Given that:
\(\log_{10} 8 = 0.90\)

\(\implies \log_{10} (2 \times 2 \times 2) = 0.90\)

\(\implies \log_{10} 2^3 = 0.90\)

\(\implies 3 \log_{10} 2 = 0.90\)

\(\implies \log_{10} 2 = \frac{0.90}{3}\)

\(\implies \log_{10} 2 = 0.30\) .... (1)

(i) To find the value of \(\log 4\):
\(\log 4 = \log_{10} (2 \times 2)\)

\(\implies = \log_{10} (2^2)\)

\(\implies = 2\log_{10} 2\)

\(\implies = 2(0.30)\) [Using equation (1)]

\(\implies = 0.60\)

(ii) To find the value of \(\log \sqrt{32}\):
\(\log \sqrt{32} = \log_{10} (32)^{\frac{1}{2}}\)

\(\implies = \frac{1}{2} \log_{10} (32)\)

\(\implies = \frac{1}{2} \log_{10} (2 \times 2 \times 2 \times 2 \times 2)\)

\(\implies = \frac{1}{2} \log_{10} (2^5)\)

\(\implies = \frac{1}{2} \times 5 \log_{10} 2\)

\(\implies = \frac{1}{2} \times 5 (0.30)\) [Using equation (1)]

\(\implies = 5 \times 0.15\)

\(\implies = 0.75\)

(iii) Find the value of \( \log_{10} 0.125 \) if \( \log_{10} 2 = 0.30 \).

\( \log_{10} 0.125 = \log_{10} \frac{125}{1000} \)
\( = \log_{10} \frac{1}{8} \)
\( = \log_{10} \frac{1}{2 \times 2 \times 2} \)
\( = \log_{10} \left(\frac{1}{2^3}\right) \)
\( = \log_{10} 2^{-3} \)
\( = -3 \log_{10} 2 \)
\( = -3 \times (0.30) \) [by substituting the value of \( \log_{10} 2 \)]
\( = -0.9 \)

In simple words: First, use the given value of \(\log_{10} 8\) to calculate the value of \(\log_{10} 2\). Then, express both 4 and \(\sqrt{32}\) as powers of 2 to easily find their values.

Exam Tip: For logarithmic calculation problems, always break down larger numbers to base 2 or 3 as early as possible so you can substitute standard known values.

Question 2. If \( \log 27 = 1.431 \), find the value of:
(i) \( \log 9 \)
(ii) \( \log 300 \)
Answer:
We are given that:
\( \log 27 = 1.431 \)
\( \implies \log 3^3 = 1.431 \)
\( \implies 3 \log 3 = 1.431 \)
\( \implies \log 3 = \frac{1.431}{3} \)
\( \implies \log 3 = 0.477 \) .....(1)

(i) Let us find \( \log 9 \):
\( \log 9 = \log 3^2 \)
\( \implies \log 9 = 2 \log 3 \)
\( \implies \log 9 = 2 \times 0.477 \) [from (1)]
\( \implies \log 9 = 0.954 \)

(ii) Let us find \( \log 300 \):
\( \log 300 = \log (3 \times 100) \)
\( \implies \log 300 = \log 3 + \log 100 \)
\( \implies \log 300 = \log 3 + 2 \) [using \( \log_{10} 100 = 2 \)]
\( \implies \log 300 = 0.477 + 2 \) [by substituting the value from (1)]
\( \implies \log 300 = 2.477 \)
In simple words: First, find the log value of 3 by using the given value of log 27. Then, write 9 and 300 in terms of 3 to solve both parts.

Exam Tip: Remember that \( \log 100 = 2 \) under standard base 10, which allows you to split and solve products containing multiples of 10 easily.

Question 3. If \( \log_{10} a = b \), find \( 10^{3b - 2} \) in terms of \( a \).
Answer:
It is given that:
\( \log_{10} a = b \)
\( \implies 10^b = a \)
Taking the cube on both sides:
\( \implies (10^b)^3 = a^3 \)
\( \implies 10^{3b} = a^3 \)
Now, divide both sides of the equation by \( 10^2 \):
\( \implies \frac{10^{3b}}{10^2} = \frac{a^3}{10^2} \)
\( \implies 10^{3b - 2} = \frac{a^3}{100} \)
In simple words: Convert the logarithm to exponential form, raise it to the third power, and then divide by 100 to get the required expression.

Exam Tip: Subtracting 2 in the exponent of a base-10 number is equivalent to dividing the value by 100.

Question 4. If \( \log_5 x = y \), find \( 5^{2y + 3} \) in terms of \( x \).
Answer:
We are given that:
\( \log_5 x = y \)
\( \implies 5^y = x \)
Squaring both sides:
\( \implies (5^y)^2 = x^2 \)
\( \implies 5^{2y} = x^2 \)
Multiplying both sides by \( 5^3 \):
\( \implies 5^{2y} \times 5^3 = x^2 \times 5^3 \)
\( \implies 5^{2y + 3} = 125x^2 \)
In simple words: Write the log equation in index form, square both sides to find \( 5^{2y} \), and then multiply by 125 to obtain the final expression.

Exam Tip: Multiplying by \( 5^3 \) (which is 125) adds 3 to the exponent on the left side, matching the required power.

Question 5. Given that \( \log_3 m = x \) and \( \log_3 n = y \), write down:
(i) \( 3^{2x - 3} \) in terms of \( m \).
(ii) \( 3^{1 - 2y + 3x} \) in terms of \( m \) and \( n \).
(iii) \( A \) in terms of \( m \) and \( n \), if \( 2 \log_3 A = 5x - 3y \).
Answer:
From the given information:
\( \log_3 m = x \)
\( \implies 3^x = m \)
\( \log_3 n = y \)
\( \implies 3^y = n \)

(i) We examine the first expression:
\( 3^{2x - 3} = 3^{2x} \cdot 3^{-3} \)
\( = 3^{2x} \cdot \frac{1}{3^3} \)
\( = \frac{(3^x)^2}{27} \)
\( = \frac{m^2}{27} \)
Consequently, \( 3^{2x - 3} = \frac{m^2}{27} \)

(ii) Let us look at the second expression:
\( 3^{1 - 2y + 3x} = 3^1 \cdot 3^{-2y} \cdot 3^{3x} \)
\( = 3 \cdot \frac{1}{3^{2y}} \cdot 3^{3x} \)
\( = \frac{3}{(3^y)^2} \cdot (3^x)^3 \)
\( = \frac{3}{n^2} \cdot m^3 \)
\( = \frac{3m^3}{n^2} \)
Consequently, \( 3^{1 - 2y + 3x} = \frac{3m^3}{n^2} \)

(iii) Now consider the third equation:
\( 2 \log_3 A = 5x - 3y \)
Substituting \( x = \log_3 m \) and \( y = \log_3 n \):
\( \implies 2 \log_3 A = 5 \log_3 m - 3 \log_3 n \)
\( \implies \log_3 A^2 = \log_3 m^5 - \log_3 n^3 \)
\( \implies \log_3 A^2 = \log_3 \left(\frac{m^5}{n^3}\right) \)
Taking the antilog on both sides:
\( \implies A^2 = \frac{m^5}{n^3} \)
Taking the square root:
\( \implies A = \sqrt{\frac{m^5}{n^3}} \)
In simple words: Convert the base logarithmic relations to exponential form first, then substitute these base forms into each expression and simplify.

Exam Tip: For multi-variable logarithmic expressions, clearly convert logarithms to base-exponent relations before substituting.

Question 6. Simplify the following:
(i) \( \log a^3 - \log a \)
(ii) \( \log a^3 \div \log a \)
Answer:
(i) \( \log a^3 - \log a = 3 \log a - \log a \)
\( = 2 \log a \)

(ii) \( \log a^3 \div \log a = \frac{3 \log a}{\log a} \)
\( = 3 \)
In simple words: Convert \( \log a^3 \) to \( 3 \log a \) using log properties, then subtract in the first part and divide in the second part.

Exam Tip: Do not confuse \( \log a^3 \) with \( (\log a)^3 \); only the former allows the exponent 3 to be brought to the front.

Question 7. If \( \log(a+b) = \log a + \log b \), find \( a \) in terms of \( b \).
Answer:
We have:
\( \log(a+b) = \log a + \log b \)
\( \implies \log(a+b) = \log ab \)
Taking the antilog on both sides:
\( \implies a + b = ab \)
Let us isolate the terms with \( a \):
\( \implies a - ab = -b \)
Multiply the equation by -1:
\( \implies ab - a = b \)
Factor out \( a \):
\( \implies a(b - 1) = b \)
\( \implies a = \frac{b}{b - 1} \)
In simple words: Combine the logarithm sum on the right side into a product, remove log from both sides, and then rearrange the equation to solve for \( a \).

Exam Tip: When expressing one variable in terms of another, group all terms containing that variable on one side and factor it out.

Question 8. Solve the following:
(i) Prove that \( (\log a)^2 - (\log b)^2 = \log(ab) \log \left(\frac{a}{b}\right) \).
(ii) If \( a \log b + b \log a - 1 = 0 \), find the value of \( a^b \cdot b^a \).
Answer:
(i) Let us verify the identity:
Left-Hand Side (L.H.S.) \( = (\log a)^2 - (\log b)^2 \)
Using the algebraic identity \( x^2 - y^2 = (x + y)(x - y) \):
\( \implies \text{L.H.S.} = (\log a + \log b)(\log a - \log b) \)
Using logarithmic laws:
\( \implies \text{L.H.S.} = \log(ab) \log \left(\frac{a}{b}\right) \)
\( \implies \text{L.H.S.} = \text{R.H.S.} \)
Thus, the identity is proved.

(ii) We are given:
\( a \log b + b \log a - 1 = 0 \)
\( \implies a \log b + b \log a = 1 \)
Using power rules:
\( \implies \log b^a + \log a^b = 1 \)
Using product rules:
\( \implies \log(b^a \cdot a^b) = \log 10 \)
Taking antilog on both sides:
\( \implies b^a \cdot a^b = 10 \)
In simple words: Use the difference of squares identity for the first proof. For the second, bring coefficients to the exponent, combine the logs, and remove the log by equating both sides.

Exam Tip: Remember to write \( 1 \) as \( \log 10 \) to easily eliminate logarithms from both sides of an equation.

Question 9. Solve the following:
(i) If \( \log(a+1) = \log(4a - 3) - \log 3 \), find the value of \( a \).
(ii) If \( 2 \log y - \log x - 3 = 0 \), express \( x \) in terms of \( y \).
(iii) Prove that \( \log_{10} 125 = 3(1 - \log_{10} 2) \).
Answer:
(i) We start with the given equation:
\( \log(a+1) = \log(4a - 3) - \log 3 \)
\( \implies \log(a+1) = \log \left(\frac{4a - 3}{3}\right) \)
By removing logs:
\( \implies a + 1 = \frac{4a - 3}{3} \)
\( \implies 3a + 3 = 4a - 3 \)
\( \implies 4a - 3a = 3 + 3 \)
\( \implies a = 6 \)

(ii) Let's rearrange the equation:
\( 2 \log y - \log x - 3 = 0 \)
\( \implies \log y^2 - \log x = 3 \)
\( \implies \log \left(\frac{y^2}{x}\right) = \log 1000 \) [since \( 3 = \log 1000 \)]
\( \implies \frac{y^2}{x} = 1000 \)
\( \implies x = \frac{y^2}{1000} \)

(iii) To prove \( \log_{10} 125 = 3(1 - \log_{10} 2) \):
Left-Hand Side (L.H.S.) \( = \log_{10} 125 \)
\( = \log_{10} 5^3 \)
\( = 3 \log_{10} 5 \) .....(1)
Right-Hand Side (R.H.S.) \( = 3(1 - \log_{10} 2) \)
\( = 3(\log_{10} 10 - \log_{10} 2) \)
\( = 3 \log_{10} \left(\frac{10}{2}\right) \)
\( = 3 \log_{10} 5 \) .....(2)
Comparing equations (1) and (2):
\( \text{L.H.S.} = \text{R.H.S.} \)
This completes the proof.
In simple words: Combine or expand terms using the quotient, product, and exponent rules of logarithms, then solve or compare the results.

Exam Tip: Convert numbers like 3 into \( \log 1000 \) or 1 into \( \log 10 \) to keep base-10 log equations uniform.

Question 10. Given \( \log x = 2m - n \), \( \log y = n - 2m \) and \( \log z = 3m - 2n \), find the value of \( \log \frac{x^2 y^3}{z^4} \) in terms of \( m \) and \( n \).
Answer:
Let us expand the expression using the properties of logarithms:
\( \log \frac{x^2 y^3}{z^4} = \log(x^2 y^3) - \log z^4 \)
\( = \log x^2 + \log y^3 - \log z^4 \)
\( = 2 \log x + 3 \log y - 4 \log z \)
Now, substitute the expressions for \( \log x \), \( \log y \), and \( \log z \):
\( = 2(2m - n) + 3(n - 2m) - 4(3m - 2n) \)
\( = 4m - 2n + 3n - 6m - 12m + 8n \)
Grouping like terms:
\( = (4 - 6 - 12)m + (-2 + 3 + 8)n \)
\( = -14m + 9n \)
In simple words: Break down the fraction log into simpler parts, bring down the powers, and then substitute the given values of log x, log y, and log z to simplify.

Exam Tip: Be careful with the minus sign when expanding \( -4(3m - 2n) \), as it becomes \( -12m + 8n \).

Question 11. Solve for \( x \): \( \log_x 25 - \log_x 5 = 2 - \log_x \frac{1}{125} \).
Answer:
Let us simplify the equation using properties of indices and logs:
\( \log_x 25 - \log_x 5 = 2 - \log_x \frac{1}{125} \)
\( \implies \log_x 5^2 - \log_x 5 = 2 - \log_x 5^{-3} \)
\( \implies 2 \log_x 5 - \log_x 5 = 2 - (-3 \log_x 5) \)
\( \implies \log_x 5 = 2 + 3 \log_x 5 \)
Rearranging the terms:
\( \implies \log_x 5 - 3 \log_x 5 = 2 \)
\( \implies -2 \log_x 5 = 2 \)
\( \implies \log_x 5 = -1 \)
Converting to index form:
\( \implies x^{-1} = 5 \)
\( \implies \frac{1}{x} = 5 \)
\( \implies x = \frac{1}{5} \)
In simple words: Convert the powers inside the logarithms to base 5, simplify each log term, collect them on one side, and solve for \( x \).

Exam Tip: Make sure to convert fractions like \( \frac{1}{125} \) to negative exponent forms like \( 5^{-3} \) to easily apply logarithmic power rules.

Exercise 8(D)

Question 1. If \( \frac{3}{2} \log a + \frac{2}{3} \log b - 1 = 0 \), find the value of \( a^9 \cdot b^4 \).
Answer:
We start with the equation:
\( \frac{3}{2} \log a + \frac{2}{3} \log b - 1 = 0 \)
\( \implies \log a^{\frac{3}{2}} + \log b^{\frac{2}{3}} = 1 \)
Combining the logarithms:
\( \implies \log \left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right) = 1 \)
Since base 10 is assumed, we write 1 as \( \log 10 \):
\( \implies \log \left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right) = \log 10 \)
\( \implies a^{\frac{3}{2}} \times b^{\frac{2}{3}} = 10 \)
Now raise both sides to the power of 6 to eliminate the fractional exponents:
\( \implies \left(a^{\frac{3}{2}} \times b^{\frac{2}{3}}\right)^6 = 10^6 \)
\( \implies a^{9} \cdot b^{4} = 10^6 \)
In simple words: Rewrite the logarithmic terms with exponents, combine them, remove the log, and raise the entire equation to the 6th power to clear the fractional exponents.

Exam Tip: The power of 6 is chosen because 6 is the Lowest Common Multiple (LCM) of the denominators of the fractional powers, 2 and 3.

Question 2. If \( x = 1 + \log 2 - \log 5 \), \( y = 2 \log 3 \) and \( z = \log a - \log 5 \); find the value of \( a \) if \( x + y = 2z \).
Answer:
Let us first simplify each of the given variables:
\( x = 1 + \log 2 - \log 5 \)
\( = \log 10 + \log 2 - \log 5 \)
\( = \log(10 \times 2) - \log 5 \)
\( = \log 20 - \log 5 \)
\( = \log \left(\frac{20}{5}\right) \)
\( = \log 4 \) .....(1)

\( y = 2 \log 3 \)
\( = \log 3^2 \)
\( = \log 9 \) .....(2)

\( z = \log a - \log 5 \)
\( = \log \left(\frac{a}{5}\right) .....(3) \)

We are given the relation:
\( x + y = 2z \)
Substitute the simplified forms from equations (1), (2), and (3):
\( \implies \log 4 + \log 9 = 2 \log \left(\frac{a}{5}\right) \)
\( \implies \log(4 \times 9) = \log \left(\frac{a}{5}\right)^2 \)
\( \implies \log 36 = \log \left(\frac{a^2}{25}\right) \)
By equating the arguments:
\( \implies \frac{a^2}{25} = 36 \)
\( \implies a^2 = 36 \times 25 \)
\( \implies a^2 = 900 \)
Taking the positive square root:
\( \implies a = 30 \)
In simple words: Simplify the given log expressions for x, y, and z first, then plug them into the equation \( x + y = 2z \) to solve for \( a \).

Exam Tip: Make sure to fully simplify individual variables before substituting them into a longer algebraic equation to prevent errors.

Question 3. If \( x = \log 0.6 \), \( y = \log 1.25 \) and \( z = \log 3 - 2 \log 2 \); find the value of:
(i) \( x + y - z \)
(ii) \( 5^{x + y - z} \)
Answer:
First, let us simplify the expression for \( z \):
\( z = \log 3 - 2 \log 2 \)
\( = \log 3 - \log 2^2 \)
\( = \log 3 - \log 4 \)
\( = \log \left(\frac{3}{4}\right) \)
\( = \log 0.75 \) .....(1)

(i) Now we compute the value of \( x + y - z \):
\( x + y - z = \log 0.6 + \log 1.25 - \log 0.75 \)
\( = \log \left(\frac{0.6 \times 1.25}{0.75}\right) \)
\( = \log \left(\frac{0.75}{0.75}\right) \)
\( = \log 1 \)
\( = 0 \) .....(2)

(ii) We now evaluate the second expression:
\( 5^{x + y - z} = 5^0 \) [using equation (2)]
\( = 1 \)
In simple words: Simplify the log terms first, add and subtract them to find that the total sum equals zero. Since any non-zero number raised to the power of zero is one, the final answer is one.

Exam Tip: Always verify that \( 0.6 \times 1.25 = 0.75 \) to correctly simplify the fraction inside the log.

Question 4. If \( a^2 = \log x \), \( b^3 = \log y \) and \( 3a^2 - 2b^3 = 6 \log z \), express \( y \) in terms of \( x \) and \( z \).
Answer:
We are given:
\( 3a^2 - 2b^3 = 6 \log z \)
Substitute \( a^2 = \log x \) and \( b^3 = \log y \):
\( \implies 3 \log x - 2 \log y = 6 \log z \)
\( \implies \log x^3 - \log y^2 = \log z^6 \)
\( \implies \log \left(\frac{x^3}{y^2}\right) = \log z^6 \)
Taking antilog on both sides:
\( \implies \frac{x^3}{y^2} = z^6 \)
Rearranging the equation to solve for \( y^2 \):
\( \implies y^2 = \frac{x^3}{z^6} \)
Taking the square root:
\( \implies y = \left(\frac{x^3}{z^6}\right)^{\frac{1}{2}} \)
\( \implies y = \frac{x^{\frac{3}{2}}}{z^3} \)
In simple words: Substitute the given values of \( a^2 \) and \( b^3 \) into the equation, combine the log terms, remove the logarithms, and solve for \( y \).

Exam Tip: Carefully apply fractional exponents when taking the square root, making sure to divide each exponent by 2.

Question 5. If \( \log \left(\frac{a - b}{2}\right) = \frac{1}{2}(\log a + \log b) \), prove that \( a^2 + b^2 = 6ab \).
Answer:
We are given:
\( \log \left(\frac{a - b}{2}\right) = \frac{1}{2}(\log a + \log b) \)
\( \implies \log \left(\frac{a - b}{2}\right) = \frac{1}{2}(\log ab) \)
\( \implies \log \left(\frac{a - b}{2}\right) = \log (ab)^{\frac{1}{2}} \)
Equating the logarithmic arguments:
\( \implies \frac{a - b}{2} = (ab)^{\frac{1}{2}} \)
Squaring both sides of the equation:
\( \implies \left(\frac{a - b}{2}\right)^2 = ab \)
\( \implies \frac{(a - b)^2}{4} = ab \)
\( \implies (a - b)^2 = 4ab \)
Expanding the left side:
\( \implies a^2 + b^2 - 2ab = 4ab \)
\( \implies a^2 + b^2 = 4ab + 2ab \)
\( \implies a^2 + b^2 = 6ab \)
Thus, the relation is verified.
In simple words: Combine the right side log terms, remove logs, square both sides to clear the square root, and expand the quadratic identity to reach the required expression.

Exam Tip: Remember that expanding \( (a-b)^2 \) gives \( a^2 - 2ab + b^2 \), which is different from \( a^2 - b^2 \).

Question 6. If \( a^2 + b^2 = 23ab \), prove that \( \log \left( \frac{a + b}{5} \right) = \frac{1}{2} (\log a + \log b) \).
Answer:
It is given that:
\( a^2 + b^2 = 23ab \)
Add \( 2ab \) to both sides of the equation:

\( \implies a^2 + b^2 + 2ab = 23ab + 2ab \)

\( \implies a^2 + b^2 + 2ab = 25ab \)

\( \implies (a + b)^2 = 25ab \)

\( \implies \frac{(a + b)^2}{25} = ab \)

\( \implies \left( \frac{a + b}{5} \right)^2 = ab \)
Take the logarithm of both sides:

\( \implies \log \left( \frac{a + b}{5} \right)^2 = \log (ab) \)

\( \implies 2 \log \left( \frac{a + b}{5} \right) = \log a + \log b \)

\( \implies \log \left( \frac{a + b}{5} \right) = \frac{1}{2} (\log a + \log b) \)
In simple words: By adding \( 2ab \) to both sides, we create a perfect square \( (a+b)^2 \). Dividing by 25 and taking logarithms on both sides simplifies the expression into the required proof.

Exam Tip: Adding \( 2ab \) to construct a perfect square is a standard algebraic technique that simplifies this derivation. Always write down each logarithm rule you use in the margin.

Question 7. If \( m = \log 20 \) and \( n = \log 25 \), find the value of \( x \), so that: \( 2 \log(x - 4) = 2m - n \).
Answer:
We are given:
\( m = \log 20 \) and \( n = \log 25 \)
The equation is:
\( 2 \log(x - 4) = 2m - n \)
Substitute the values of \( m \) and \( n \) into the equation:

\( \implies 2 \log(x - 4) = 2 \log 20 - \log 25 \)

\( \implies \log (x - 4)^2 = \log (20^2) - \log 25 \)

\( \implies \log (x - 4)^2 = \log 400 - \log 25 \)

\( \implies \log (x - 4)^2 = \log \left( \frac{400}{25} \right) \)

\( \implies \log (x - 4)^2 = \log 16 \)
Since the logarithms on both sides have the same base:

\( \implies (x - 4)^2 = 16 \)
Taking the square root on both sides:

\( \implies x - 4 = 4 \) (since the base or argument of a logarithm cannot be negative, we discard the negative root)

\( \implies x = 4 + 4 \)

\( \implies x = 8 \)
In simple words: First replace \( m \) and \( n \) with the given values. Use the laws of subtraction and powers to combine the logarithms on the right side. Taking the square root of both sides gives \( x = 8 \).

Exam Tip: Be careful to discard the negative root when removing square exponents, because logarithms are only defined for positive real numbers.

Question 8. Solve for \( x \) and \( y \); if \( x > 0 \) and \( y > 0 \):
\( \log(xy) = \log\left(\frac{x}{y}\right) + 2 \log 2 = 2 \)
Answer:
First, split the compound equation into two separate equations:
\( \log(xy) = 2 \)
Convert this into exponential form with base 10:

\( \implies xy = 10^2 \)

\( \implies xy = 100 \dots (1) \)
Now take the second equation:
\( \log\left(\frac{x}{y}\right) + 2 \log 2 = 2 \)
Using log properties:

\( \implies \log\left(\frac{x}{y}\right) + \log (2^2) = 2 \log 10 \)

\( \implies \log\left(\frac{x}{y}\right) + \log 4 = \log (10^2) \)

\( \implies \log\left(\frac{x}{y}\right) + \log 4 = \log 100 \)
Combine the terms on the left using the product rule:

\( \implies \log \left( \frac{x}{y} \times 4 \right) = \log 100 \)
Removing logarithms from both sides:

\( \implies \frac{4x}{y} = 100 \)

\( \implies 4x = 100y \)

\( \implies x = 25y \)
Substitute this expression for \( x \) into equation (1):

\( \implies (25y) \times y = 100 \)

\( \implies 25y^2 = 100 \)

\( \implies y^2 = \frac{100}{25} \)

\( \implies y^2 = 4 \)
Since it is given that \( y > 0 \), we take only the positive root:

\( \implies y = 2 \)
Substitute \( y = 2 \) back into equation (1) to get \( x \):

\( \implies x \times 2 = 100 \)

\( \implies x = \frac{100}{2} \)

\( \implies x = 50 \)
Thus, the solutions are \( x = 50 \) and \( y = 2 \).
In simple words: Split the system into two distinct equations. Solve the first to find \( xy = 100 \), and combine the second to find \( x = 25y \). Substitute \( x \) to find \( y = 2 \) and \( x = 50 \).

Exam Tip: Remember that any base-less logarithm in ICSE is assumed to have a base of 10. Always convert them to exponential forms to make solving simultaneous equations simpler.

Question 9. Find \( x \), if:
(i) \( \log_x 625 = -4 \)
(ii) \( \log_x (5x - 6) = 2 \)
Answer:
(i) Given equation:
\( \log_x 625 = -4 \)
Convert the equation to exponential form:

\( \implies 625 = x^{-4} \)

\( \implies 5^4 = \left( \frac{1}{x} \right)^4 \)
Since the exponents on both sides are the same, equate the bases:

\( \implies 5 = \frac{1}{x} \)

\( \implies x = \frac{1}{5} \)

(ii) Given equation:
\( \log_x (5x - 6) = 2 \)
Convert to exponential form:

\( \implies 5x - 6 = x^2 \)
Rearranging this as a standard quadratic equation:

\( \implies x^2 - 5x + 6 = 0 \)
Factorise the quadratic expression:

\( \implies x^2 - 3x - 2x + 6 = 0 \)

\( \implies x(x - 3) - 2(x - 3) = 0 \)

\( \implies (x - 2)(x - 3) = 0 \)
This gives the roots:

\( \therefore x = 2 \text{ or } x = 3 \)
In simple words: For part (i), convert the logarithm into an index form and solve. For part (ii), rewrite the logarithm as a quadratic equation and factorise to get two values for \( x \).

Exam Tip: For logarithmic bases, always double-check if your solutions are greater than 0 and not equal to 1. Since both 2 and 3 satisfy this, both are valid.

Question 10. If \( p = \log 20 \) and \( q = \log 25 \), find the value of \( x \), if \( 2 \log(x + 1) = 2p - q \).
Answer:
We are given:
\( p = \log 20 \) and \( q = \log 25 \)
The equation to solve is:
\( 2 \log(x + 1) = 2p - q \)
Substitute the values of \( p \) and \( q \) into this equation:

\( \implies 2 \log(x + 1) = 2 \log 20 - \log 25 \)
Using logarithm power rules:

\( \implies \log (x + 1)^2 = \log (20^2) - \log 25 \)

\( \implies \log (x + 1)^2 = \log 400 - \log 25 \)
Apply the quotient rule to simplify the subtraction on the right:

\( \implies \log (x + 1)^2 = \log \left( \frac{400}{25} \right) \)

\( \implies \log (x + 1)^2 = \log 16 \)
Express 16 as a perfect square:

\( \implies \log (x + 1)^2 = \log (4^2) \)
Equate the arguments since the logarithm bases are identical:

\( \implies (x + 1)^2 = 4^2 \)
Since \( x + 1 \) must be positive for the log to be defined, we take the positive square root:

\( \implies x + 1 = 4 \)

\( \implies x = 4 - 1 \)

\( \implies x = 3 \)
In simple words: Put the log values in for \( p \) and \( q \). Apply logarithm rules to condense the right side down to \( \log 16 \), which lets you set up \( (x+1)^2 = 16 \) and solve for \( x \).

Exam Tip: Never expand the term \( (x+1)^2 \) unless necessary. Writing both sides of the equation as perfect squares makes finding the square root straightforward.

Question 11. If \( \log_2(x + y) = \log_3(x - y) = \frac{\log 25}{\log 0.2} \), find the values of \( x \) and \( y \).
Answer:
Let us first simplify the constant term on the right:
\( \frac{\log 25}{\log 0.2} = \log_{0.2} 25 \)
Since \( 0.2 = \frac{2}{10} = \frac{1}{5} = 5^{-1} \) and \( 25 = 5^2 \):

\( \implies \log_{0.2} 25 = \log_{5^{-1}} (5^2) \)

\( \implies \log_{0.2} 25 = \frac{2}{-1} \log_5 5 \)

\( \implies \log_{0.2} 25 = -2 \)
Now, set the first logarithmic term equal to \( -2 \):
\( \log_2 (x + y) = -2 \)
Convert to exponential form:

\( \implies x + y = 2^{-2} \)

\( \implies x + y = \frac{1}{4} \dots (i) \)
Next, set the second logarithmic term equal to \( -2 \):
\( \log_3 (x - y) = -2 \)
Convert to exponential form:

\( \implies x - y = 3^{-2} \)

\( \implies x - y = \frac{1}{9} \dots (ii) \)
We can find \( x \) and \( y \) by solving equations (i) and (ii). Add the two equations:

\( \implies (x + y) + (x - y) = \frac{1}{4} + \frac{1}{9} \)

\( \implies 2x = \frac{9 + 4}{36} \)

\( \implies 2x = \frac{13}{36} \)

\( \implies x = \frac{13}{72} \)
Now, subtract equation (ii) from equation (i):

\( \implies (x + y) - (x - y) = \frac{1}{4} - \frac{1}{9} \)

\( \implies 2y = \frac{9 - 4}{36} \)

\( \implies 2y = \frac{5}{36} \)

\( \implies y = \frac{5}{72} \)
Thus, the values are \( x = \frac{13}{72} \) and \( y = \frac{5}{72} \).
In simple words: Simplify the fraction on the right to \( -2 \). This gives you two basic equations: \( x + y = \frac{1}{4} \) and \( x - y = \frac{1}{9} \), which can be solved easily for \( x \) and \( y \).

Exam Tip: Mastering the base-change rule \( \frac{\log a}{\log b} = \log_b a \) is crucial for simplifying complex multi-base problems.

Question 12. Given \( \frac{\log x}{\log y} = \frac{3}{2} \) and \( \log(xy) = 5 \); find the values of \( x \) and \( y \).
Answer:
From the first given equation:
\( \frac{\log x}{\log y} = \frac{3}{2} \)

\( \implies 2 \log x = 3 \log y \)

\( \implies \log y = \frac{2 \log x}{3} \dots (i) \)
Now consider the second equation:
\( \log(xy) = 5 \)
Apply the log product rule:

\( \implies \log x + \log y = 5 \)
Substitute equation (i) into this relation:

\( \implies \log x + \frac{2 \log x}{3} = 5 \)

\( \implies \frac{3 \log x + 2 \log x}{3} = 5 \)

\( \implies \frac{5 \log x}{3} = 5 \)

\( \implies \log x = 3 \)
Convert to exponential base 10 form:

\( \implies x = 10^3 \)

\( \therefore x = 1000 \)
Substitute \( \log x = 3 \) back into equation (i):

\( \implies \log y = \frac{2 \times 3}{3} \)

\( \implies \log y = 2 \)
Convert to exponential base 10 form:

\( \implies y = 10^2 \)

\( \therefore y = 100 \)
In simple words: Express \( \log y \) in terms of \( \log x \) using the first equation, and substitute it into the second equation. This gives \( x = 1000 \) and \( y = 100 \).

Exam Tip: Ensure that you do not confuse \( \log(xy) \) with \( \log x \times \log y \). The product rule expands \( \log(xy) \) to \( \log x + \log y \).

Question 13. Given \( \log_{10} x = 2a \) and \( \log_{10} y = \frac{b}{2} \).
(i) Write \( 10^a \) in terms of \( x \).
(ii) Write \( 10^{2b+1} \) in terms of \( y \).
(iii) If \( \log_{10} P = 3a - 2b \), express \( P \) in terms of \( x \) and \( y \).
Answer:
(i) We are given:
\( \log_{10} x = 2a \)
Convert this into exponential form:

\( \implies x = 10^{2a} \)
Take the square root of both sides:

\( \implies x^{1/2} = (10^{2a})^{1/2} \)

\( \implies x^{1/2} = 10^a \)

\( \implies 10^a = x^{1/2} \)

(ii) We are given:
\( \log_{10} y = \frac{b}{2} \)
Convert this into exponential form:

\( \implies y = 10^{b/2} \)
Raise both sides to the power of 4:

\( \implies y^4 = (10^{b/2})^4 \)

\( \implies y^4 = 10^{2b} \)
Multiply both sides by 10:

\( \implies 10 y^4 = 10^{2b} \times 10 \)

\( \implies 10^{2b+1} = 10y^4 \)

(iii) Given relation:
\( \log_{10} P = 3a - 2b \)
Convert to exponential form:

\( \implies P = 10^{3a - 2b} \)
Use the rules of indices to expand:

\( \implies P = \frac{10^{3a}}{10^{2b}} \)

\( \implies P = \frac{(10^a)^3}{(10^b)^2} \)
We know from part (i) that \( 10^a = x^{1/2} \). Also, since \( y = 10^{b/2} \), squaring both sides gives \( 10^b = y^2 \). Substitute these in:

\( \implies P = \frac{(x^{1/2})^3}{(y^2)^2} \)

\( \implies P = \frac{x^{3/2}}{y^4} \)
In simple words: Convert the logs to power forms first. Use algebraic rules for exponents to write \( 10^a \), \( 10^{2b+1} \), and \( P \) in terms of \( x \) and \( y \).

Exam Tip: Be meticulous with fractional power rules such as \( (10^{b/2})^4 = 10^{2b} \) and index division rules to keep your algebra correct.

Question 14. Solve: \( \log_5(x + 1) - 1 = 1 + \log_5(x - 1) \).
Answer:
The given equation is:
\( \log_5(x + 1) - 1 = 1 + \log_5(x - 1) \)
Rearrange the terms to group the logarithms on the left-hand side:

\( \implies \log_5(x + 1) - \log_5(x - 1) = 1 + 1 \)

\( \implies \log_5(x + 1) - \log_5(x - 1) = 2 \)
Apply the quotient law of logarithms:

\( \implies \log_5 \left( \frac{x+1}{x-1} \right) = 2 \)
Convert to exponential form:

\( \implies \frac{x+1}{x-1} = 5^2 \)

\( \implies \frac{x+1}{x-1} = 25 \)
Cross-multiply to solve for \( x \):

\( \implies x + 1 = 25(x - 1) \)

\( \implies x + 1 = 25x - 25 \)

\( \implies 25x - x = 25 + 1 \)

\( \implies 24x = 26 \)

\( \implies x = \frac{26}{24} = \frac{13}{12} \)
In simple words: Move both log terms to one side of the equation and combine them using division. Turn it into a basic linear fraction to get \( x = \frac{13}{12} \).

Exam Tip: Always make sure to write the final fraction in its simplest reduced form to secure full marks.

Question 15. Solve for \( x \): \( \log_x 49 - \log_x 7 + \log_x \frac{1}{343} = -2 \).
Answer:
The given expression is:
\( \log_x 49 - \log_x 7 + \log_x \frac{1}{343} = -2 \)
Combine the logarithms using product and division properties:

\( \implies \log_x \left( \frac{49}{7 \times 343} \right) = -2 \)
Simplify the fraction inside the argument:

\( \implies \log_x \left( \frac{1}{49} \right) = -2 \)
This can be written as:

\( \implies -\log_x 49 = -2 \)

\( \implies \log_x 49 = 2 \)
Convert to exponential form:

\( \implies 49 = x^2 \)
Taking the square root on both sides (since base \( x > 0 \)):

\( \therefore x = 7 \)
In simple words: Combine all log terms on the left using division. This simplifies the equation to \( \log_x (1/49) = -2 \), which gives \( x^2 = 49 \) or \( x = 7 \).

Exam Tip: A logarithm base must always be positive and cannot equal 1. Thus, the negative root \( x = -7 \) is discarded.

Question 16. If \( a^2 = \log x \), \( b^3 = \log y \) and \( \frac{a^2}{2} - \frac{b^3}{3} = \log c \); find \( c \) in terms of \( x \) and \( y \).
Answer:
We are given that:
\( a^2 = \log x \) and \( b^3 = \log y \)
The relation is:
\( \frac{a^2}{2} - \frac{b^3}{3} = \log c \)
Substitute \( a^2 \) and \( b^3 \) into this relation:

\( \implies \frac{\log x}{2} - \frac{\log y}{3} = \log c \)
Find a common denominator on the left side:

\( \implies \frac{3 \log x - 2 \log y}{6} = \log c \)
Cross-multiply:

\( \implies 3 \log x - 2 \log y = 6 \log c \)
Apply power laws for logarithms:

\( \implies \log x^3 - \log y^2 = \log c^6 \)
Combine using the quotient law:

\( \implies \log \left( \frac{x^3}{y^2} \right) = \log c^6 \)
Remove logarithms from both sides:

\( \implies \frac{x^3}{y^2} = c^6 \)
Take the sixth root of both sides:

\( \implies c = \sqrt[6]{\frac{x^3}{y^2}} \)
In simple words: Plug the given values into the equation, merge the logarithms on the left into one term using quotient laws, and then solve for \( c \).

Exam Tip: Moving coefficients inside the logarithm as exponents (e.g. \( 3 \log x = \log x^3 \)) is a standard method that must be performed before subtraction is applied.

Question 17. Given \( x = \log_{10} 12 \), \( y = \log_4 2 \times \log_{10} 9 \), and \( z = \log_{10} 0.4 \), find the value of:
(i) \( x - y - z \)
(ii) \( 13^{x - y - z} \)
Answer:
(i) Let us find the value of \( x - y - z \) by substituting the given values:
\( x - y - z = \log_{10} 12 - \log_4 2 \times \log_{10} 9 - \log_{10} 0.4 \)
Using logarithm properties, rewrite the individual terms:

\( = \log_{10} (4 \times 3) - \log_4 2 \times \log_{10} (3^2) - \log_{10} 0.4 \)

\( = \log_{10} 4 + \log_{10} 3 - \log_{2^2} 2 \times 2 \log_{10} 3 - \log_{10} \left( \frac{4}{10} \right) \)

\( = \log_{10} 4 + \log_{10} 3 - \frac{1}{2} \log_2 2 \times 2 \log_{10} 3 - (\log_{10} 4 - \log_{10} 10) \)
Since \( \log_2 2 = 1 \) and \( \log_{10} 10 = 1 \):

\( = \log_{10} 4 + \log_{10} 3 - \log_{10} 3 - \log_{10} 4 + 1 \)
Cancel out matching terms:

\( = 1 \)

(ii) Using the value calculated in part (i):
\( 13^{x-y-z} = 13^1 = 13 \)
In simple words: Expand and split each log into prime bases. After cancellation, part (i) simplifies to 1. For part (ii), 13 to the power of 1 is 13.

Exam Tip: Be comfortable with changing bases such as \( \log_4 2 = \log_{2^2} 2 = \frac{1}{2} \). These patterns frequently appear in exam questions.

Question 18. Solve for \( x \): \( \log_x 15\sqrt{5} = 2 - \log_x 3\sqrt{5} \).
Answer:
The given equation is:
\( \log_x 15\sqrt{5} = 2 - \log_x 3\sqrt{5} \)
Move the log terms together on the left side of the equation:

\( \implies \log_x 15\sqrt{5} + \log_x 3\sqrt{5} = 2 \)
Apply the product law of logarithms:

\( \implies \log_x \left( 15\sqrt{5} \times 3\sqrt{5} \right) = 2 \)

\( \implies \log_x (45 \times 5) = 2 \)

\( \implies \log_x 225 = 2 \)
Write 225 as a perfect square:

\( \implies \log_x (15^2) = 2 \)

\( \implies 2 \log_x 15 = 2 \)
Divide both sides by 2:

\( \implies \log_x 15 = 1 \)
Convert this into exponential form:

\( \implies x^1 = 15 \)

\( \implies x = 15 \)
In simple words: Combine the log terms on the left using the multiplication rule. This simplifies the expression to \( \log_x 225 = 2 \), which easily solves to \( x = 15 \).

Exam Tip: Remember to multiply the square roots properly: \( \sqrt{5} \times \sqrt{5} = 5 \), which helps in cleanly calculating the product \( 15 \times 3 \times 5 = 225 \).

Question 19. Evaluate:
(i) \( \log_b a \times \log_c b \times \log_a c \)
(ii) \( \log_3 8 \div \log_9 16 \)
(iii) \( \frac{\log_5 8}{\log_{25} 16 \times \log_{100} 10} \)
Answer:
(i) Change the bases to a common base (like base 10):
\( \log_b a \times \log_c b \times \log_a c = \frac{\log a}{\log b} \times \frac{\log b}{\log c} \times \frac{\log c}{\log a} \)
All numerator and denominator terms cancel out:

\( = 1 \)

(ii) Write the expression as a fraction:
\( \log_3 8 \div \log_9 16 = \frac{\log_3 8}{\log_9 16} \)
Apply the base-change formula:

\( = \frac{\frac{\log 8}{\log 3}}{\frac{\log 16}{\log 9}} \)

\( = \frac{\log 8}{\log 3} \times \frac{\log 9}{\log 16} \)
Rewrite the arguments as prime powers:

\( = \frac{\log (2^3)}{\log 3} \times \frac{\log (3^2)}{\log (2^4)} \)

\( = \frac{3 \log 2}{\log 3} \times \frac{2 \log 3}{4 \log 2} \)
After canceling \( \log 2 \) and \( \log 3 \):

\( = \frac{3 \times 2}{4} = \frac{6}{4} = \frac{3}{2} \)

(iii) Given fraction:
\( \frac{\log_5 8}{\log_{25} 16 \times \log_{100} 10} \)
Convert each logarithm using the base-change formula:

\( = \frac{\frac{\log 8}{\log 5}}{\frac{\log 16}{\log 25} \times \frac{\log 10}{\log 100}} \)
Express terms as powers of primes (2, 5, 10):

\( = \frac{\frac{\log (2^3)}{\log 5}}{\frac{\log (2^4)}{\log (5^2)} \times \frac{\log 10}{\log (10^2)}} \)
Rearranging the fractional division:

\( = \frac{\log (2^3)}{\log 5} \times \frac{\log (5^2)}{\log (2^4)} \times \frac{\log (10^2)}{\log 10} \)
Apply power rule on logarithms:

\( = \frac{3 \log 2}{\log 5} \times \frac{2 \log 5}{4 \log 2} \times \frac{2 \log 10}{\log 10} \)
Cancel out common logarithmic factors:

\( = \frac{3 \times 2 \times 2}{4} = \frac{12}{4} = 3 \)
In simple words: Rewrite all log terms using the base-change rule so that they are in common bases. Express the arguments as prime powers to cancel terms out cleanly.

Exam Tip: Converting division of fractions into multiplication by the reciprocal is a standard algebraic step that helps prevent errors in multi-tier fraction problems.

Question 20. Evaluate: \( \log_a m \div \log_{ab} m \).
Answer:
The division is written as:
\( \log_a m \div \log_{ab} m = \frac{\log_a m}{\log_{ab} m} \)
Use the reciprocal rule \( \log_y x = \frac{1}{\log_x y} \) to swap base and argument:

\( = \frac{\frac{1}{\log_m a}}{\frac{1}{\log_m ab}} \)

\( = \frac{\log_m (ab)}{\log_m a} \)
Apply the base-change rule:

\( = \log_a (ab) \)
Apply the log product rule:

\( = \log_a a + \log_a b \)
Since \( \log_a a = 1 \):

\( = 1 + \log_a b \)
In simple words: Use the base swap rule to write both logs with the same base \( m \). This reduces the expression to \( \log_a (ab) \), which expands to \( 1 + \log_a b \).

Exam Tip: When different bases share the same argument, swapping bases using the reciprocal rule simplifies the expression dramatically.