ICSE Class 9 Maths Chapter 08 Logarithms

8. Logarithms

8.1 Introduction

Logarithms are used to make the long and complicated calculations easy. Consider \(3^4 = 81\), this is the exponential form of representing relation between three numbers 3, 4 and 81. Now the same relation between 3, 4 and 81 can be written as

\(\log_3 81 = 4\) (read as: logarithm of 81 at base 3 is 4).

Thus:

\(3^4 = 81 \Leftrightarrow \log_3 81 = 4\)

Definition: If a, b and c are three real numbers such that \(a \neq 1\) and \(a^b = c\) then b is called logarithm of c at the base a and is written as \(\log_a c = b\); read as log of c at the base a is b.

\(a^b = c \Leftrightarrow \log_a c = b\)

Teacher's Note

Logarithms were historically used by scientists and engineers to perform complex calculations before calculators were invented. Today, they remain essential in fields like seismology, where the Richter scale measures earthquake magnitude using logarithmic formulas.

8.2 Interchanging (Logarithmic Form Vis-a-Vis Exponential Form)

\(a^b = c\) is called the exponential form

and, \(\log_a c = b\) is called the logarithmic form.

i.e., (i) \(2^{-3} = 0.125\) (Exponential form)

\(\Rightarrow\) log of 0.125 to the base 2 = -3

i.e., \(\log_2 0.125 = -3\) (Logarithmic form)

(ii) \(\log_{64} 8 = \frac{1}{2}\) (Logarithmic form)

\(\Rightarrow\) log of 8 to the base 64 = \(\frac{1}{2}\)

i.e. \((64)^{\frac{1}{2}} = 8\) (Exponential form) and so on.

Similarly:

If x is positive;

(iii) \(x^0 = 1 \Rightarrow \log_x 1 = 0\) i.e. log of 1 to the base x = 0

In general; the logarithm of 1 to any base is zero.

i.e. \(\log_x 1 = 0\); \(\log_{10} 1 = 0\); \(\log_a 1 = 0\) and so on.

(iv) \(x^1 = x \Rightarrow \log_x x = 1\) i.e. log x to the base x = 1

In general; the logarithm of any number to the same base is always one.

i.e. \(\log_5 5 = 1\); \(\log_{10} 10 = 1\); \(\log_a a = 1\) and so on.

Teacher's Note

Understanding the relationship between exponential and logarithmic forms helps students recognize that logarithms are simply the inverse of exponentiation, making complex calculations more manageable.

Example 1

Find: (i) the logarithm of 1000 to the base 10.

(ii) the logarithm of \(\frac{1}{9}\) to the base 3.

Solution:

(i) Let \(\log_{10} 1000 = x\) \(\Rightarrow 10^x = 1000\)

\(\Rightarrow 10^x = 10^3 \Rightarrow x = 3\)

\(\therefore \log_{10} 1000 = 3\) Ans.

(ii) Let \(\log_3 \frac{1}{9} = x\) \(\Rightarrow 3^x = \frac{1}{9}\)

\(\Rightarrow 3^x = 3^{-2} \Rightarrow x = -2\)

\(\therefore \log_3 \frac{1}{9} = -2\) Ans.

Example 2

Find x, if: (i) \(\log_2 x = -2\) (ii) \(\log_4 (x + 3) = 2\) (iii) \(\log_x 64 = \frac{3}{2}\)

Solution:

(i) \(\log_2 x = -2\) \(\Rightarrow 2^{-2} = x\)

\(\Rightarrow x = \frac{1}{4}\) Ans.

(ii) \(\log_4 (x + 3) = 2\) \(\Rightarrow 4^2 = x + 3\)

\(\Rightarrow x = 16 - 3 = 13\) Ans.

(iii) \(\log_x 64 = \frac{3}{2}\) \(\Rightarrow x^{\frac{3}{2}} = 64\)

\(\Rightarrow x = (64)^{\frac{2}{3}} = (2^6)^{\frac{2}{3}} = 2^4 = 16\) Ans.

Teacher's Note

Converting between logarithmic and exponential forms is a fundamental skill that helps students solve real-world problems in finance, medicine, and environmental science where growth and decay follow logarithmic patterns.

Exercise 8(A)

1. Express each of the following in logarithmic form:

(i) \(5^3 = 125\) (ii) \(3^{-2} = \frac{1}{9}\)

(iii) \(10^{-3} = 0.001\) (iv) \((81)^{\frac{3}{4}} = 27\)

(vi) \(\frac{1}{16}\) to the base 4

(vii) 27 to the base 9

(viii) \(\frac{1}{81}\) to the base 27

2. Express each of the following in exponential form:

(i) \(\log_8 0.125 = -1\) (ii) \(\log_{10} 0.01 = -2\)

(iii) \(\log_a A = x\) (iv) \(\log_{10} 1 = 0\)

3. Solve for x: \(\log_{10} x = -2\).

4. Find the logarithm of:

(i) 100 to the base 10

(ii) 0.1 to the base 10

(iii) 0.001 to the base 10

(iv) 32 to the base 4

(v) 0.125 to the base 2

5. State, true or false:

(i) If \(\log_{10} x = a\), then \(10^x = a\).

(ii) If \(x^y = z\), then \(y = \log_z x\).

(iii) \(\log_2 8 = 3\) and \(\log_8 2 = \frac{1}{3}\).

6. Find x, if:

(i) \(\log_3 x = 0\) (ii) \(\log_5 2 = -1\)

(iii) \(\log_9 243 = x\) (iv) \(\log_5 (x - 7) = 1\)

(v) \(\log_4 32 = x - 4\) (vi) \(\log_9 (2x^2 - 1) = 2\)

7. Evaluate:

(i) \(\log_{10} 0.01\) (ii) \(\log_2 (1 + 8)\)

(iii) \(\log_3 1\) (iv) \(\log_5 125\)

(v) \(\log_{16} 8\) (vi) \(\log_{0.5} 16\)

8. If \(\log_a m = n\), express \(a^{n-1}\) in terms of a and m.

\(\log_a m = n \Rightarrow a^n = m\)

\(\Rightarrow a^{n-1} = \frac{a^n}{a} = \frac{m}{a}\)

9. Given \(\log_x m = m\) and \(\log_x y = n\).

(i) Express \(2^{m-3}\) in terms of x.

(ii) Express \(5^{3n+2}\) in terms of y.

10. If \(\log_x x = a\) and \(\log_y y = a\), write \(72^a\) in terms of x and y.

11. Solve for x: \(\log(x - 1) + \log(x + 1) = \log_3 1\).

12. If \(\log(x^2 - 21) = 2\), show that \(x = \pm 11\).

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