RBSE Solutions Class 11 Maths Chapter 9 Logarithms Exercise 9.2

Question 1. Prove that: \( \log 630 = \log 2 + 2 \log 3 + \log 5 + \log 7 \).
Answer: We need to show that the left side equals the right side. Let's start with the left-hand side (L.H.S.).
L.H.S. \( = \log 630 \)
First, factorize 630 into its prime factors. \( 630 = 2 \times 3 \times 3 \times 5 \times 7 \)
\( = \log (2 \times 3 \times 3 \times 5 \times 7) \)
\( = \log (2 \times 3^2 \times 5 \times 7) \)
Using the logarithm property \( \log(abc) = \log a + \log b + \log c \):
\( = \log 2 + \log 3^2 + \log 5 + \log 7 \)
Then, using the property \( \log a^b = b \log a \):
\( = \log 2 + 2 \log 3 + \log 5 + \log 7 \)
This matches the right-hand side (R.H.S.) of the equation. So, the statement is proven.
In simple words: We broke down 630 into its smaller prime numbers. Then, we used logarithm rules to separate these numbers with plus signs and brought the power of 3 (which is 2) to the front. This made both sides of the equation the same.

🎯 Exam Tip: When proving logarithmic identities, always start with one side (usually the more complex one) and simplify it using logarithm properties until it matches the other side. Factorization is key for numbers.

Question 2. Prove that: \( \log \frac{9}{14} + \log \frac{35}{24} - \log \frac{15}{16} = 0 \).
Answer: We need to show that the left side simplifies to 0. Let's start with the left-hand side (L.H.S.).
\[ -\log \frac{a}{b} = \log \frac{b}{a} \]
L.H.S. \( = \log \frac{9}{14} + \log \frac{35}{24} - \log \frac{15}{16} \)
First, apply the logarithm property \( \log a - \log b = \log (a/b) \) or \( \log a + \log b = \log (a \times b) \). We can change the minus sign to a plus sign by flipping the fraction.
\( = \log \frac{9}{14} + \log \frac{35}{24} + \log \frac{16}{15} \)
Now, use the property \( \log a + \log b + \log c = \log (a \times b \times c) \).
\( = \log \left( \frac{9}{14} \times \frac{35}{24} \times \frac{16}{15} \right) \)
Next, factorize the numbers to simplify the multiplication.
\( = \log \left( \frac{3 \times 3}{2 \times 7} \times \frac{5 \times 7}{3 \times 8} \times \frac{2 \times 8}{3 \times 5} \right) \)
Now, cancel out common factors from the numerator and denominator.
\( = \log \left( \frac{\cancel{3} \times \cancel{3} \times \cancel{5} \times \cancel{7} \times \cancel{2} \times \cancel{8}}{\cancel{2} \times \cancel{7} \times \cancel{3} \times \cancel{8} \times \cancel{3} \times \cancel{5}} \right) \)
After cancelling all factors, we are left with 1.
\( = \log (1) \)
We know that \( \log 1 = 0 \).
\( = 0 \)
This matches the right-hand side (R.H.S.) of the equation. So, the statement is proven.
In simple words: We combined all the logarithm terms into a single logarithm by multiplying the fractions inside. We changed the minus sign to a plus by flipping the last fraction. After multiplying, all the numbers cancelled out, leaving us with log 1, which is always 0.

🎯 Exam Tip: Remember to simplify fractions by cancelling common factors before multiplying, especially in logarithm problems. The key properties \( \log a + \log b = \log (ab) \) and \( \log a - \log b = \log (a/b) \) are crucial.

Question 3. Prove that: \( \log 10 + \log 100 + \log 1000 + \log 10000 = 10 \).
Answer: We need to show that the left side equals the right side. Let's start with the left-hand side (L.H.S.).
L.H.S. \( = \log 10 + \log 100 + \log 1000 + \log 10000 \)
First, express each number as a power of 10.
\( = \log 10^1 + \log 10^2 + \log 10^3 + \log 10^4 \)
Next, use the logarithm property \( \log a^b = b \log a \).
\( = 1 \log 10 + 2 \log 10 + 3 \log 10 + 4 \log 10 \)
Now, factor out \( \log 10 \).
\( = (1 + 2 + 3 + 4) \log 10 \)
\( = 10 \log 10 \)
Since the base of the logarithm is not specified, it's usually assumed to be 10 (common logarithm). So, \( \log 10 = 1 \).
\( = 10 \times 1 \)
\( = 10 \)
This matches the right-hand side (R.H.S.) of the equation. So, the statement is proven.
In simple words: We changed 100, 1000, and 10000 into powers of 10. Then, we moved the powers to the front of the log. Adding up all the numbers in front gave us 10, and because \( \log 10 \) is 1, the answer is 10.

🎯 Exam Tip: Remember that \( \log_{10} 10 = 1 \) (or simply \( \log 10 = 1 \) if the base is implied). Expressing numbers as powers of the base simplifies calculations significantly.

Question 4. If \( \log 2 = 0.3010 \), \( \log 3 = 0.4771 \), \( \log 7 = 0.8451 \) and \( \log 11 = 1.0414 \), then find the value of the following :
(i) \( \log 36 \)
(ii) \( \log \frac{42}{11} \)
(iii) \( \log \left(\frac{11}{7}\right)^5 \)
(iv) \( \log 70 \)
(v) \( \log \frac{121}{120} \)
(vi) \( \log 5^{1/3} \)
Answer:
(i) To find \( \log 36 \):
First, express 36 as a product of its prime factors.
\( \log 36 = \log (2 \times 2 \times 3 \times 3) \)
\( = \log (2^2 \times 3^2) \)
Using the property \( \log (ab) = \log a + \log b \):
\( = \log 2^2 + \log 3^2 \)
Using the property \( \log a^b = b \log a \):
\( = 2 \log 2 + 2 \log 3 \)
Substitute the given values for \( \log 2 \) and \( \log 3 \).
\( = 2(0.3010) + 2(0.4771) \)
\( = 0.6020 + 0.9542 \)
\( = 1.5562 \)

(ii) To find \( \log \frac{42}{11} \):
Using the property \( \log (a/b) = \log a - \log b \):
\( \log \frac{42}{11} = \log 42 - \log 11 \)
First, express 42 as a product of its prime factors.
\( = \log (2 \times 3 \times 7) - \log 11 \)
Using the property \( \log (abc) = \log a + \log b + \log c \):
\( = \log 2 + \log 3 + \log 7 - \log 11 \)
Substitute the given values for \( \log 2, \log 3, \log 7 \), and \( \log 11 \).
\( = 0.3010 + 0.4771 + 0.8451 - 1.0414 \)
\( = 1.6232 - 1.0414 \)
\( = 0.5818 \)

(iii) To find \( \log \left(\frac{11}{7}\right)^5 \):
Using the property \( \log a^b = b \log a \):
\( \log \left(\frac{11}{7}\right)^5 = 5 \log \frac{11}{7} \)
Using the property \( \log (a/b) = \log a - \log b \):
\( = 5(\log 11 - \log 7) \)
Substitute the given values for \( \log 11 \) and \( \log 7 \).
\( = 5(1.0414 - 0.8451) \)
\( = 5(0.1963) \)
\( = 0.9815 \)

(iv) To find \( \log 70 \):
Express 70 as a product of numbers whose logarithms are known or can be found.
\( \log 70 = \log (7 \times 10) \)
Using the property \( \log (ab) = \log a + \log b \):
\( = \log 7 + \log 10 \)
Substitute the given value for \( \log 7 \) and use \( \log 10 = 1 \).
\( = 0.8451 + 1 \)
\( = 1.8451 \)

(v) To find \( \log \frac{121}{120} \):
Using the property \( \log (a/b) = \log a - \log b \):
\( \log \frac{121}{120} = \log 121 - \log 120 \)
Express 121 and 120 as products of their prime factors.
\( = \log 11^2 - \log (12 \times 10) \)
Using the property \( \log a^b = b \log a \) and \( \log (ab) = \log a + \log b \):
\( = 2 \log 11 - (\log 12 + \log 10) \)
Further factorize 12.
\( = 2 \log 11 - (\log (2 \times 2 \times 3) + \log 10) \)
\( = 2 \log 11 - (\log 2^2 + \log 3 + \log 10) \)
\( = 2 \log 11 - (2 \log 2 + \log 3 + \log 10) \)
Substitute the given values: \( \log 11 = 1.0414 \), \( \log 2 = 0.3010 \), \( \log 3 = 0.4771 \), \( \log 10 = 1 \).
\( = 2(1.0414) - (2(0.3010) + 0.4771 + 1) \)
\( = 2.0828 - (0.6020 + 0.4771 + 1) \)
\( = 2.0828 - (2.0791) \)
\( = 0.0037 \)
In simple words: For each part, we first broke down the numbers into prime factors like 2, 3, 7, and 11, or used 10. Then we used the basic rules of logarithms (like log of multiplication is adding logs, log of division is subtracting logs, and log of a power means multiplying the power) to rewrite the expressions. Finally, we put in the given decimal values and calculated the answer.

🎯 Exam Tip: Always break down numbers into their prime factors first, especially when dealing with multiplication or division inside logarithms. Remember the fundamental properties: \( \log(ab) = \log a + \log b \), \( \log(a/b) = \log a - \log b \), and \( \log a^b = b \log a \).

Question 5. Find the value of x from following equation \( \log_x 4 + \log_x 16 + \log_x 64 = 12 \).
Answer: We need to solve for x in the given equation.
The equation is: \( \log_x 4 + \log_x 16 + \log_x 64 = 12 \)
First, express 16 and 64 as powers of 4.
\( \implies \log_x 4 + \log_x 4^2 + \log_x 4^3 = 12 \)
Using the logarithm property \( \log_b a^c = c \log_b a \):
\( \implies \log_x 4 + 2 \log_x 4 + 3 \log_x 4 = 12 \)
Now, combine the terms with \( \log_x 4 \).
\( \implies (1 + 2 + 3) \log_x 4 = 12 \)
\( \implies 6 \log_x 4 = 12 \)
Divide both sides by 6.
\( \implies \log_x 4 = \frac{12}{6} \)
\( \implies \log_x 4 = 2 \)
Now, convert the logarithmic equation to an exponential equation. If \( \log_b a = c \), then \( b^c = a \).
\( \implies x^2 = 4 \)
Since 4 can be written as \( 2^2 \), we have:
\( \implies x^2 = 2^2 \)
On comparing the bases, we find x.
\( \implies x = 2 \)
In simple words: We rewrote 16 and 64 as powers of 4. Then we used a log rule to bring the powers to the front. This allowed us to add all the terms with \( \log_x 4 \). After simplifying, we changed the log equation into a power equation to find x.

🎯 Exam Tip: When all terms have the same base and you need to find the base (x), try to express the numbers as powers of a common factor. Remember to convert the final logarithmic equation into an exponential form to solve for x.

Question 6. Solve the equation : \( \log (x + 1) - \log (x - 1) = 1 \).
Answer: We need to solve for x in the given equation.
The equation is: \( \log (x + 1) - \log (x - 1) = 1 \)
Using the logarithm property \( \log a - \log b = \log (a/b) \):
\( \implies \log \left( \frac{x+1}{x-1} \right) = 1 \)
Since the base of the logarithm is not specified, it is assumed to be 10 (common logarithm). So, we can write 1 as \( \log 10 \).
\( \implies \log \left( \frac{x+1}{x-1} \right) = \log 10 \)
Now, if \( \log A = \log B \), then \( A = B \).
\( \implies \frac{x+1}{x-1} = 10 \)
Multiply both sides by \( (x-1) \).
\( \implies x+1 = 10(x-1) \)
Distribute 10 on the right side.
\( \implies x+1 = 10x - 10 \)
Move all x terms to one side and constant terms to the other side.
\( \implies 1 + 10 = 10x - x \)
\( \implies 11 = 9x \)
Divide by 9 to find x.
\( \implies x = \frac{11}{9} \)
In simple words: We combined the two log terms into one by dividing the expressions inside. Since the right side was 1, we wrote it as \( \log 10 \). Then, we removed the logs from both sides and solved the simple equation to find the value of x.

🎯 Exam Tip: When solving logarithmic equations, first use logarithm properties to combine multiple log terms into a single log term. If the base is not given, assume it's 10. Remember to convert to exponential form or equate the arguments if both sides are in log form.

Question 8. Give the solution of following questions in one term :
(i) \( \log 2 + 1 \)
(ii) \( \log 2x + 2 \log x \)
Answer:
(i) To simplify \( \log 2 + 1 \):
Since the base of the logarithm is not specified, it's usually assumed to be 10. We know that \( \log 10 = 1 \).
\( = \log 2 + \log 10 \)
Using the logarithm property \( \log a + \log b = \log (ab) \):
\( = \log (2 \times 10) \)
\( = \log 20 \)

(ii) To simplify \( \log 2x + 2 \log x \):
Using the logarithm property \( b \log a = \log a^b \):
\( = \log 2x + \log x^2 \)
Using the logarithm property \( \log a + \log b = \log (ab) \):
\( = \log (2x \times x^2) \)
\( = \log (2x^3) \)
In simple words: For the first part, we changed the number 1 into \( \log 10 \) and then combined it with \( \log 2 \) by multiplying the numbers inside. For the second part, we moved the 2 in front of \( \log x \) to become a power of x, then combined both log terms by multiplying their contents.

🎯 Exam Tip: When simplifying expressions, remember that '1' can be written as \( \log_{b} b \). Also, always look for opportunities to apply the power rule of logarithms to combine terms effectively.

Question 9. Prove that:
(i) \( \log_5 3 \cdot \log_3 4 \cdot \log_2 5 = 2 \)
(ii) \( \log_a x \times \log_b y = \log_b x \times \log_a y \).
Answer:
(i) To prove \( \log_5 3 \cdot \log_3 4 \cdot \log_2 5 = 2 \):
L.H.S. \( = \log_5 3 \cdot \log_3 4 \cdot \log_2 5 \)
Using the change of base formula \( \log_b a = \frac{\log_c a}{\log_c b} \), we can change all to base 10 (or any common base).
\[ \log_b n = \frac{\log_a n}{\log_a b} \]
\( = \frac{\log_{10} 3}{\log_{10} 5} \times \frac{\log_{10} 4}{\log_{10} 3} \times \frac{\log_{10} 5}{\log_{10} 2} \)
Now, cancel out common terms from the numerator and denominator.
\( = \frac{\cancel{\log_{10} 3}}{\cancel{\log_{10} 5}} \times \frac{\log_{10} 4}{\cancel{\log_{10} 3}} \times \frac{\cancel{\log_{10} 5}}{\log_{10} 2} \)
\( = \frac{\log_{10} 4}{\log_{10} 2} \)
We know that \( 4 = 2^2 \).
\( = \frac{\log_{10} 2^2}{\log_{10} 2} \)
Using the property \( \log a^b = b \log a \):
\( = \frac{2 \log_{10} 2}{\log_{10} 2} \)
Cancel \( \log_{10} 2 \).
\( = 2 \)
This matches the right-hand side (R.H.S.). So, the statement is proven.

(ii) To prove \( \log_a x \times \log_b y = \log_b x \times \log_a y \):
L.H.S. \( = \log_a x \times \log_b y \)
Using the change of base formula \( \log_b a = \frac{\log_c a}{\log_c b} \), change to a common base (e.g., base 10).
\[ \log_b n = \frac{\log_a n}{\log_a b} \]
\( = \frac{\log_{10} x}{\log_{10} a} \times \frac{\log_{10} y}{\log_{10} b} \)
Now, rearrange the terms in the multiplication.
\( = \frac{\log_{10} y}{\log_{10} a} \times \frac{\log_{10} x}{\log_{10} b} \)
Using the change of base formula again, but in reverse ( \( \frac{\log_c a}{\log_c b} = \log_b a \) ):
\( = \log_a y \times \log_b x \)
This can also be written as \( \log_b x \times \log_a y \), which matches the right-hand side (R.H.S.). So, the statement is proven.
In simple words: For the first part, we used a rule to change all the logs to a common base, which allowed many terms to cancel out. Then we simplified the remaining part by turning 4 into \( 2^2 \) and used another rule to bring the power down, resulting in 2. For the second part, we also used the base change rule for all logs. After rewriting them with a common base, we just rearranged the multiplication to match the other side.

🎯 Exam Tip: The change of base formula is fundamental for simplifying and proving logarithmic identities. When you have products of logarithms with different bases, converting them to a common base often reveals cancellations or rearrangements that lead to the solution.