OP Malhotra Class 9 Maths Solutions Chapter 7 Logarithms Exercise 7 (A)

Question 1. Given an equivalent exponential form for each statement :
(i) log2 8 = 3
(ii) log3 81 = 4
(iii) log2 \( \frac { 1 }{ 2 } \) = − 1
(iv) log5 \( \frac { 1 }{ 25 } \)= − 2
(v) log4 2 = \( \frac { 1 }{ 2 } \)
(vi) log27 3 = \( \frac { 1 }{ 3 } \)
Answer:
(i) \( \log_2 8 = 3 \implies 2^3 = 8 \)
(ii) \( \log_3 81 = 4 \implies 3^4 = 81 \)
(iii) \( \log_2 \frac { 1 }{ 2 } = -1 \implies 2^{-1} = \frac { 1 }{ 2 } \)
(iv) \( \log_5 \frac { 1 }{ 25 } = -2 \implies 5^{-2} = \frac { 1 }{ 25 } \)
(v) \( \log_4 2 = \frac { 1 }{ 2 } \implies 4^{\frac { 1 }{ 2 }} = 2 \) (This means the square root of 4 is 2.)
(vi) \( \log_{27} 3 = \frac { 1 }{ 3 } \implies 27^{\frac { 1 }{ 3 }} = 3 \)
In simple words: Logarithms and exponentials are two ways of saying the same thing. If you have \( \log_b a = x \), it simply means that \( b \) raised to the power of \( x \) gives you \( a \). We are just rewriting each statement using this rule.

🎯 Exam Tip: Remember the core definition: \( \log_b a = x \) is equivalent to \( b^x = a \). Clearly identify the base, the number, and the logarithm to easily convert between forms.

Question 2. Give an equivalent logarithmic form for each statement :
(i) 16 = 24
(ii) 25 = 52
(iii) 81 = 34
(iv) 6° = 1
(v) 8\( \frac { 1 }{ 3 } \) = 2
(vi) \( \frac { 1 }{ 9 } \) = 3-2
(vii) \( \frac {1}{32} \) = 2-5
(viii) 101.4969 = 31.4
Answer:
(i) \( 16 = 2^4 \implies \log_2 16 = 4 \)
(ii) \( 25 = 5^2 \implies \log_5 25 = 2 \)
(iii) \( 81 = 3^4 \implies \log_3 81 = 4 \)
(iv) \( 6^0 = 1 \implies \log_6 1 = 0 \) (Any non-zero number raised to the power of zero is one.)
(v) \( 8^{\frac { 1 }{ 3 }} = 2 \implies \log_8 2 = \frac { 1 }{ 3 } \)
(vi) \( \frac { 1 }{ 9 } = 3^{-2} \implies \log_3 \frac { 1 }{ 9 } = -2 \)
(vii) \( \frac { 1 }{ 32 } = 2^{-5} \implies \log_2 \frac { 1 }{ 32 } = -5 \)
(viii) \( 10^{1.4969} = 31.4 \implies \log_{10} 31.4 = 1.4969 \)
In simple words: We are doing the opposite of the last question. If you have a number \( b \) raised to the power of \( x \) that gives \( a \) (like \( b^x = a \)), you can write it as \( \log_b a = x \). Just identify the base, the exponent, and the result.

🎯 Exam Tip: Pay close attention to the base of the logarithm. It's the number that has the exponent in the exponential form.

Question 3. Find the value of each logarithm given below:
(i) log10 1000
(ii) log2 8
(iii) log3 81
(iv) log10 0.1
(v) log10 0.01
(vi) log10 0.0001
(vii) log2 \( \frac { 1 }{ 4 } \)
(viii) log3 \( \frac { 1 }{ 27 } \)
(ix) log3 1
(x) log\( _{\frac{1}{2}} \frac{1}{4} \)
(xi) log27 9
(xii) log\( _{\frac{1}{125}} 125 \)
(xiii) log\( _{\frac{5}{6}} 1 \)
(xiv) log\( _{\frac{1}{3}} 9 \)
(xv) log10 10
Answer:
(i) Let \( \log_{10} 1000 = x \).
\( \implies 10^x = 1000 \)
\( \implies 10^x = 10^3 \)
Comparing the exponents, we get \( x = 3 \).
Therefore, \( \log_{10} 1000 = 3 \).

(ii) Let \( \log_2 8 = x \).
\( \implies 2^x = 8 \)
\( \implies 2^x = 2^3 \)
Comparing the exponents, we get \( x = 3 \).
Therefore, \( \log_2 8 = 3 \).

(iii) Let \( \log_3 81 = x \).
\( \implies 3^x = 81 \)
\( \implies 3^x = 3^4 \)
Comparing the exponents, we get \( x = 4 \).
Therefore, \( \log_3 81 = 4 \).

(iv) Let \( \log_{10} 0.1 = x \).
\( \implies 10^x = 0.1 \)
\( \implies 10^x = \frac{1}{10} \)
\( \implies 10^x = 10^{-1} \)
Comparing the exponents, we get \( x = -1 \).
Therefore, \( \log_{10} 0.1 = -1 \).

(v) Let \( \log_{10} 0.01 = x \).
\( \implies 10^x = 0.01 \)
\( \implies 10^x = \frac{1}{100} \)
\( \implies 10^x = \frac{1}{10^2} \)
\( \implies 10^x = 10^{-2} \)
Comparing the exponents, we get \( x = -2 \).
Therefore, \( \log_{10} 0.01 = -2 \).

(vi) Let \( \log_{10} 0.0001 = x \).
\( \implies 10^x = 0.0001 \)
\( \implies 10^x = \frac{1}{10000} \)
\( \implies 10^x = \frac{1}{10^4} \)
\( \implies 10^x = 10^{-4} \)
Comparing the exponents, we get \( x = -4 \).
Therefore, \( \log_{10} 0.0001 = -4 \).

(vii) Let \( \log_2 \frac { 1 }{ 4 } = x \).
\( \implies 2^x = \frac { 1 }{ 4 } \)
\( \implies 2^x = \frac { 1 }{ 2^2 } \)
\( \implies 2^x = 2^{-2} \)
Comparing the exponents, we get \( x = -2 \).
Therefore, \( \log_2 \frac { 1 }{ 4 } = -2 \).

(viii) Let \( \log_3 \frac { 1 }{ 27 } = x \).
\( \implies 3^x = \frac{1}{27} \)
\( \implies 3^x = \frac{1}{3^3} \)
\( \implies 3^x = 3^{-3} \)
Comparing the exponents, we get \( x = -3 \).
Therefore, \( \log_3 \frac { 1 }{ 27 } = -3 \).

(ix) Let \( \log_3 1 = x \).
\( \implies 3^x = 1 \)
\( \implies 3^x = 3^0 \) (Remember that any number (except 0) raised to the power of 0 is 1.)
Comparing the exponents, we get \( x = 0 \).
Therefore, \( \log_3 1 = 0 \).

(x) Let \( \log_{\frac{1}{2}} \frac{1}{4} = x \).
\( \implies \left(\frac{1}{2}\right)^x = \frac{1}{4} \)
\( \implies \left(\frac{1}{2}\right)^x = \left(\frac{1}{2}\right)^2 \)
Comparing the exponents, we get \( x = 2 \).
Therefore, \( \log_{\frac{1}{2}} \frac{1}{4} = 2 \).

(xi) Let \( \log_{27} 9 = x \).
\( \implies 27^x = 9 \)
\( \implies (3^3)^x = 3^2 \)
\( \implies 3^{3x} = 3^2 \)
Comparing the exponents, we get \( 3x = 2 \).
\( \implies x = \frac{2}{3} \).
Therefore, \( \log_{27} 9 = \frac{2}{3} \).

(xii) Let \( \log_{\frac{1}{125}} 125 = x \).
\( \implies \left(\frac{1}{125}\right)^x = 125 \)
\( \implies (125^{-1})^x = 125^1 \)
\( \implies 125^{-x} = 125^1 \)
Comparing the exponents, we get \( -x = 1 \).
\( \implies x = -1 \).
Therefore, \( \log_{\frac{1}{125}} 125 = -1 \).

(xiii) Let \( \log_{\frac{5}{6}} 1 = x \).
\( \implies \left(\frac{5}{6}\right)^x = 1 \)
\( \implies \left(\frac{5}{6}\right)^x = \left(\frac{5}{6}\right)^0 \)
Comparing the exponents, we get \( x = 0 \).
Therefore, \( \log_{\frac{5}{6}} 1 = 0 \).

(xiv) Let \( \log_{\frac{1}{3}} 9 = x \).
\( \implies \left(\frac{1}{3}\right)^x = 9 \)
\( \implies (3^{-1})^x = 3^2 \)
\( \implies 3^{-x} = 3^2 \)
Comparing the exponents, we get \( -x = 2 \).
\( \implies x = -2 \).
Therefore, \( \log_{\frac{1}{3}} 9 = -2 \).

(xv) Let \( \log_{10} 10 = x \).
\( \implies 10^x = 10 \)
\( \implies 10^x = 10^1 \)
Comparing the exponents, we get \( x = 1 \).
Therefore, \( \log_{10} 10 = 1 \).
In simple words: To find the value of a logarithm, think: "What power do I need to raise the base to, to get the number inside the log?" Then, write both sides of the equation with the same base and match the powers to find \( x \). This method makes it easy to compare and solve for \( x \).

🎯 Exam Tip: When finding the value of a logarithm, always try to express the number inside the logarithm as a power of the base. This simplifies the equation and makes finding the exponent easier.

Question 4. Find the value of x
(i) logx 216 = 3
(ii) log4 x = -4
(iii) log3 x = 0
(iv) log8 x = \( \frac { 2 }{ 3 } \)
(v) log10 100 = x
(vi) log2 0.5 = x
Answer:
(i) Given \( \log_x 216 = 3 \).
\( \implies x^3 = 216 \)
\( \implies x^3 = 6^3 \)
Comparing the bases, we get \( x = 6 \).

(ii) Given \( \log_4 x = -4 \).
\( \implies 4^{-4} = x \)
\( \implies x = \frac{1}{4^4} \)
\( \implies x = \frac{1}{256} \).

(iii) Given \( \log_3 x = 0 \).
\( \implies 3^0 = x \)
\( \implies x = 1 \) (Any number raised to the power of zero is 1.)

(iv) Given \( \log_8 x = \frac{2}{3} \).
\( \implies 8^{\frac{2}{3}} = x \)
\( \implies x = (2^3)^{\frac{2}{3}} \)
\( \implies x = 2^{3 \times \frac{2}{3}} \)
\( \implies x = 2^2 \)
\( \implies x = 4 \).

(v) Given \( \log_{10} 100 = x \).
\( \implies 10^x = 100 \)
\( \implies 10^x = 10^2 \)
Comparing the exponents, we get \( x = 2 \).

(vi) Given \( \log_2 0.5 = x \).
\( \implies 2^x = 0.5 \)
\( \implies 2^x = \frac{1}{2} \)
\( \implies 2^x = 2^{-1} \)
Comparing the exponents, we get \( x = -1 \).
In simple words: To find \( x \) in these questions, change the logarithm equation into its exponential form. Then, either match the bases to compare exponents, or match the exponents to compare bases. This helps solve for \( x \) easily.

🎯 Exam Tip: When the unknown \( x \) is the base or the result of the logarithm, converting to exponential form is the most direct approach. Remember your exponent rules, especially for negative and fractional powers.

Question 5. Answer true or false : If log10x = a, then 10ª = x.
Answer: The statement is True.
This is the fundamental definition of a logarithm. If \( \log_b a = x \), it means that \( b \) raised to the power of \( x \) gives \( a \). In this case, \( b = 10 \), the number \( a \) is \( x \), and the logarithm value is \( a \). So, \( \log_{10} x = a \) is indeed equivalent to \( 10^a = x \).
In simple words: The statement is true. A logarithm simply asks, "What power do I need for this base to get that number?" So, if \( \log_{10} x = a \), it means 10 to the power of \( a \) equals \( x \).

🎯 Exam Tip: Understanding the basic definition of logarithms is crucial. \( \log_b a = x \) means \( b^x = a \). This relationship helps in solving problems and converting between logarithmic and exponential forms.