Logarithm Class 9 ICSE — Maths Chapter 8

Written by Greya LakshmiPublished on 12 October 2026

Logarithms turn long multiplication and division into simpler addition and subtraction. This page walks through Chapter 8 of the ICSE Class 9 Mathematics textbook, where the logarithm is defined, the two forms are interchanged, and the laws are applied. Here is the logarithm class 9 ICSE chapter explained section by section. You can browse the rest of the Class 9 Mathematics book hub or the full Class 9 book list.

ICSE Class 9 Mathematics Chapter 8 Logarithms PDF

What is inside Chapter 8

10 pages · 5 sections · 4 worked examples · 59 exercise questions

Section	Title
8.1	INTRODUCTION
8.2	INTERCHANGING
8.3	LAWS OF LOGARITHM WITH USE
8.4	EXPANSION OF EXPRESSIONS WITH THE HELP OF LAWS OF LOGARITHM
8.5	MORE ABOUT LOGARITHMS

EXERCISE 8(A): 12 questions
EXERCISE 8(B): 16 questions
EXERCISE 8(C): 11 questions
EXERCISE 8(D): 20 questions

8.1 INTRODUCTION

Logarithms exist to make long calculations easy. Instead of multiplying large numbers directly, you can add exponents. The chapter opens by fixing the meaning of a logarithm.

If $a$ , $b$ and $c$ are real numbers, with $a \neq 1$ , and $a^b = c$ , then $b$ is the logarithm of $c$ at base $a$ . It is written $\log_a c = b$ . The two forms are equivalent:

$a^b = c \Leftrightarrow \log_a c = b$

Worked pattern: take $2^5 = 32$ . Here $a = 2$ , $b = 5$ , $c = 32$ . The same relation is $\log_2 32 = 5$ . The base is the number being raised, not the result.

The condition $a \neq 1$ keeps the logarithm unique, because $1^x = 1$ for every exponent. The power notation here comes from the earlier chapter on indices and exponents. For the CISCE syllabus and exam details, see the official CISCE website. The later chapter on triangles continues the book.

8.2 INTERCHANGING

The same statement has two forms. $a^b = c$ is the exponential form. $\log_a c = b$ is the logarithmic form. To switch forms, keep the base in place and move the output across.

Two values appear constantly. For any positive $x$ :

$x^0 = 1$ gives $\log_x 1 = 0$ . The logarithm of 1 to any base is 0.
$x^1 = x$ gives $\log_x x = 1$ . The logarithm of a number to its own base is 1.

Worked pattern: solve $\log_2 x = -3$ . Convert to exponential form: $2^{-3} = x$ , so $x = \frac{1}{8}$ . A negative logarithm means the number is a fraction.

Second pattern: solve $\log_x 27 = \frac{3}{2}$ . Convert to $x^{\frac{3}{2}} = 27$ . Raise both sides to the power $\frac{2}{3}$ : $x = 27^{\frac{2}{3}} = (3^3)^{\frac{2}{3}} = 3^2 = 9$ .

The base in $\log_x$ cannot be 1, as stated in the definition. This section also asks you to express powers such as $10^{-3} = 0.001$ in log form and to convert logs such as $\log_{10} 0.01 = -2$ back to powers. Those conversions are the same move in reverse.

8.3 LAWS OF LOGARITHM WITH USE

The laws tell you how logarithms behave with multiplication, division and powers. All three need the same base.

First Law (Product Law): $\log_a (m \times n) = \log_a m + \log_a n$ .

Second Law (Quotient Law): $\log_a \frac{m}{n} = \log_a m - \log_a n$ .

Third Law (Power Law): $\log_a (m)^n = n \log_a m$ . A useful corollary follows from $\sqrt[n]{m} = m^{\frac{1}{n}}$ : $\log_a \sqrt[n]{m} = \frac{1}{n} \log_a m$ .

Common mistakes to avoid:

$\log_a (m + n) \neq \log_a m + \log_a n$
$\frac{\log_a m}{\log_a n} \neq \log_a m - \log_a n$
$\log_a (m - n) \neq \log_a m - \log_a n$

Logarithms to base 10 are called common logarithms. When no base is written, the base is 10: $\log 8 = \log_{10} 8$ . That is why $\log_{10} 1 = 0$ , $\log_{10} 10 = 1$ , and $\log_{10} 100 = 2$ .

Worked pattern: expand $\log_a \frac{p^3 q}{r^2}$ . First apply the quotient law: $\log_a p^3 + \log_a q - \log_a r^2$ . Then apply the power law: $3 \log_a p + \log_a q - 2 \log_a r$ .

8.4 EXPANSION OF EXPRESSIONS WITH THE HELP OF LAWS OF LOGARITHM

Expansion means taking one compact log expression and writing it as separate log terms. The order of application is steady: quotient law first, then product law, then power law.

Worked pattern: expand $y = \frac{a^3 b}{c^2}$ . Then $\log y = \log(a^3 b) - \log c^2 = \log a^3 + \log b - \log c^2 = 3\log a + \log b - 2\log c$ .

The reverse direction condenses separate terms into one logarithm. Pattern: if $\log V = \log 2 + 3\log r - \log 5$ , rewrite $3\log r$ as $\log r^3$ . Then $\log V = \log 2 + \log r^3 - \log 5 = \log \frac{2 r^3}{5}$ , so $V = \frac{2 r^3}{5}$ .

Roots follow the same rules. Worked pattern: express $\log_{10} \sqrt[3]{72}$ in terms of $\log_{10} 2$ and $\log_{10} 3$ . Since $72 = 2^3 \times 3^2$ ,

$\log_{10} \sqrt[3]{72} = \frac{1}{3} \log_{10} 72 = \frac{1}{3} (3 \log_{10} 2 + 2 \log_{10} 3).$

If $\log 2 = 0.3010$ and $\log 3 = 0.4771$ , you can evaluate $\log 12$ : $\log 12 = \log (2^2 \times 3) = 2\log 2 + \log 3 = 0.6020 + 0.4771 = 1.0791$ .

8.5 MORE ABOUT LOGARITHMS

The reciprocal rule connects two logs whose base and argument are swapped. For positive $a$ and $b$ :

$\log_b a = \frac{1}{\log_a b}, \quad \log_b a \times \log_a b = 1.$

Also, since $\log_a a = 1$ , multiplying by $x$ gives $\log_a a^x = x$ . For example, $\log_3 3^7 = 7$ .

The change of base formula lets you evaluate a log in any base you can handle. For positive $a$ , $b$ and $x$ :

$\log_b a = \frac{\log_x a}{\log_x b}.$

Worked pattern: evaluate $\log_{16} 8$ . Change to base 2: $\frac{\log_2 8}{\log_2 16} = \frac{3}{4}$ . The same method handles differences such as $\log_9 27 - \log_4 8 = \frac{3}{2} - \frac{3}{2} = 0$ .

Exam pattern note: a bare $\log 2$ in a question means $\log_{10} 2$ . The reciprocal rule then gives a fast check — $\log_2 7 \times \log_7 2 = 1$ — and the change of base formula converts awkward bases into base 10 or any working base.

This chapter belongs to the Class 9 hub, and the wider site map starts at the ICSE home page.

What to practise from this chapter

The chapter has four exercise blocks. Work through them in order.

Exercise 8(A) — 12 questions
Exercise 8(B) — 16 questions
Exercise 8(C) — 11 questions
Exercise 8(D) — 20 questions

Frequently asked questions

Why is log of 1 equal to 0 for any base?

Because $x^0=1$ for every positive $x$ , the exponent that takes any base to 1 is 0. So $\log_x 1 = 0$ .

What does it mean when no base is written, like log 8?

It means base 10. Logs to base 10 are common logarithms, so $\log 8$ means $\log_{10}8$ .

When can I split a logarithm into two separate logs?

Only when the input is a product, quotient, or power. The laws give $\log_a (m \times n) = \log_a m + \log_a n$ and $\log_a \frac{m}{n} = \log_a m - \log_a n$ . A sum such as $\log_a (m+n)$ cannot be split.

Why does log_b a times log_a b equal 1?

Because $\log_b a = \frac{1}{\log_a b}$ . The two expressions are reciprocals, so their product is 1.

The syllabus this chapter follows is set by the CISCE board.

Reference: ICSE Class 9 Mathematics textbook material, chapter 8: Logarithms — CISCE-aligned.

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