MathsICSE Class 9 · Chapter 7

Logarithms ICSE Class 9: Log Form, Laws of Logarithms and Expansion

Q: Is the change-of-base formula in the Class IX syllabus?

The syllabus lists interchanging forms and the laws of logarithms with expansion. Focus on these; change of base is not named.

Q: Will I need log tables?

The syllabus says certain questions may require mathematical tables, so learn to read them, but most Class IX questions give values such as log 2 = 0.3010.

Q: Why must the base not be 1?

Every power of 1 is 1, so log₁ N would have no single value.

Q: Can a logarithm be negative?

Yes. log 0.01 = −2 because 10⁻² = 0.01. Only the number inside the log must be positive.

Q: How are logarithms linked to indices?

A logarithm is an index. Each law of logarithms comes from a law of indices.

Updated 15 October 2026 · Fact-checked

A logarithm is an index: if aˣ = N, then logₐ N = x. ICSE Class 9 asks you to switch between exponential and logarithmic form, use the laws log mn = log m + log n, log (m/n) = log m − log n and log mⁿ = n log m, and expand expressions such as log (a⁴b²/c³).

What Logarithms is about

Logarithms are another way of writing indices. Since $2^{5} = 32$ , we say 'the logarithm of 32 to the base 2 is 5' and write $lo g_{2} 32 = 5$ . In general:

$a^{x} = N ⟺ lo g_{a} N = x$ , where $a > 0$ , $a \neq = 1$ and $N > 0$ .

The CISCE 2028 Class IX syllabus has two parts: (a) interchanging logarithmic and exponential form, and (b) the laws of logarithms and their use, including expanding an expression such as $y = \frac{a ^{4} \times b ^{2}}{c ^{3}}$ into $lo g y = 4 lo g a + 2 lo g b - 3 lo g c$ . When no base is written, the base is 10 (common logarithms). The paper may also allow logarithm tables.

School exam questions include converting forms, evaluating logs such as $lo g_{2} 8$ , finding unknowns like $x$ in $lo g_{x} 243 = 5$ , expanding and combining logs, finding values given $lo g 2$ and $lo g 3$ , and proving simple results.

Weightage: CISCE publishes no chapter weightage and no question-paper layout for ICSE Class 9 Mathematics. Class IX is examined by your school: one 80-mark paper of 3 hours plus 20 marks of internal assessment. Logarithms is in the 3. Algebra (v) part of the 2028 syllabus.Source: CISCE ICSE 2028 Mathematics syllabus, Class IX

Key concepts

Exponential form and logarithmic form

$a^{x} = N$ (exponential form) and $lo g_{a} N = x$ (logarithmic form) say exactly the same thing. Read $lo g_{a} N$ as 'the power to which $a$ must be raised to give $N$ '.

Conditions: the base $a > 0$ and $a \neq = 1$ ; the number $N > 0$ . So the logarithm of zero or a negative number is not defined.

Two results follow straight away: $lo g_{a} 1 = 0$ (since $a^{0} = 1$ ) and $lo g_{a} a = 1$ (since $a^{1} = a$ ).

Worked example

Write in logarithmic form: (i) $3^{4} = 81$ (ii) $1 0^{- 2} = 0.01$ . Write in exponential form: (iii) $lo g_{5} 125 = 3$ .

Try it first, then show the solutionSolution

(i) $lo g_{3} 81 = 4$ .
(ii) $lo g_{10} 0.01 = - 2$ .
(iii) $5^{3} = 125$ .

Answer: (i) log₃ 81 = 4 (ii) log₁₀ 0.01 = −2 (iii) 5³ = 125

Evaluating logarithms and finding unknowns

To evaluate $lo g_{a} N$ , let it equal $x$ , change to exponential form $a^{x} = N$ , write both sides as powers of the same base, and equate indices.

The same idea finds an unknown base or number:

$lo g_{x} 243 = 5$ means $x^{5} = 243 = 3^{5}$ , so $x = 3$ .
$lo g_{5} x = - 2$ means $x = 5^{- 2} = \frac{1}{25}$ .

Worked example

Evaluate $lo g_{2} 8$ .

Try it first, then show the solutionSolution

Let $lo g_{2} 8 = x$ , so $(2)^{x} = 8$ .
$2^{\frac{x}{2}} = 2^{3}$ .
$\frac{x}{2} = 3$ , so $x = 6$ .

Answer: 6

Laws of logarithms

For $m, n > 0$ and a valid base $a$ :

1. Product law: $lo g_{a} (mn) = lo g_{a} m + lo g_{a} n$ .
2. Quotient law: $lo g_{a} \frac{m}{n} = lo g_{a} m - lo g_{a} n$ .
3. Power law: $lo g_{a} m^{n} = n lo g_{a} m$ .

These come from the laws of indices: if $m = a^{p}$ and $n = a^{q}$ then $mn = a^{p + q}$ , so $lo g_{a} mn = p + q$ .

There is no law for $lo g (m + n)$ or $lo g m \times lo g n$ . Also $\frac{l o g m}{l o g n}$ is not $lo g m - lo g n$ .

Worked example

Prove the product law $lo g_{a} mn = lo g_{a} m + lo g_{a} n$ .

Try it first, then show the solutionSolution

Let $lo g_{a} m = p$ and $lo g_{a} n = q$ , so $m = a^{p}$ and $n = a^{q}$ .
$mn = a^{p} \times a^{q} = a^{p + q}$ .
In log form: $lo g_{a} mn = p + q = lo g_{a} m + lo g_{a} n$ .

Answer: Proved.

Expanding an expression

To expand, break a single log into a sum and difference of simpler logs:

products become sums;
quotients become differences;
powers and roots become multipliers (a root is a fractional power: $x = x^{\frac{1}{2}}$ ).

This is exactly the syllabus example: if $y = \frac{a ^{4} \times b ^{2}}{c ^{3}}$ , then $lo g y = 4 lo g a + 2 lo g b - 3 lo g c$ .

Worked example

Expand $lo g \frac{x ^{3} y}{z ^{2}}$ .

Try it first, then show the solutionSolution

$= \frac{1}{2} lo g \frac{x ^{3} y}{z ^{2}}$ .
$= \frac{1}{2} (3 lo g x + lo g y - 2 lo g z)$ .
$= \frac{3}{2} lo g x + \frac{1}{2} lo g y - lo g z$ .

Answer: $\frac{3}{2} lo g x + \frac{1}{2} lo g y - lo g z$

Combining into a single logarithm

Run the laws backwards:
1. Move every multiplier into the index: $2 lo g 5 = lo g 25$ .
2. Add logs by multiplying numbers; subtract by dividing.
3. Write any whole number as a log: with base 10, $1 = lo g 10$ and $2 = lo g 100$ .

For example, $1 + lo g 2 = lo g 10 + lo g 2 = lo g 20$ .

Worked example

Show that $lo g \frac{75}{16} - 2 lo g \frac{5}{9} + lo g \frac{32}{243} = lo g 2$ .

Try it first, then show the solutionSolution

$2 lo g \frac{5}{9} = lo g \frac{25}{81}$ .
Left side $= lo g (\frac{75}{16} \times \frac{81}{25} \times \frac{32}{243})$ .
$\frac{75 \times 81 \times 32}{16 \times 25 \times 243} = \frac{3 \times 2}{3} = 2$ , so the left side is $lo g 2$ .

Answer: log 2, as required.

Using given values such as log 2 and log 3

Factorise the number into primes, then use the laws. With $lo g 2 = 0.3010$ and $lo g 3 = 0.4771$ (base 10):

$lo g 12 = 2 lo g 2 + lo g 3 = 1.0791$ ;
$lo g 5 = lo g \frac{10}{2} = 1 - lo g 2 = 0.6990$ ;
$lo g 1.5 = lo g 3 - lo g 2 = 0.1761$ .

The trick $5 = \frac{10}{2}$ lets you find logs of numbers with factor 5 using only $lo g 2$ .

Worked example

Given $lo g 2 = 0.3010$ and $lo g 3 = 0.4771$ , find $lo g 72$ .

Try it first, then show the solutionSolution

$72 = 2^{3} \times 3^{2}$ .
$lo g 72 = 3 lo g 2 + 2 lo g 3 = 0.9030 + 0.9542$ .
$= 1.8572$ .

Answer: 1.8572

Maths in real life: Logarithms

Each concept, met in an everyday situation. Read the situation, try the problem, then check the solution. The one-line hook is what to remember.

Logarithm as an index (decibels)

Suppose a sound-level gauge uses the rule loudness $L = 10 lo g \frac{I}{I _{0}}$ decibels, where $I_{0}$ is the faintest sound you can hear.

Find the loudness of a sound 1,000 times as intense as $I_{0}$ , and of one 1,00,000 times as intense.

Try it first, then show the solutionSolution

$L = 10 lo g 1000 = 10 \times 3 = 30$ dB.
$L = 10 lo g 1 0^{5} = 10 \times 5 = 50$ dB.

Answer: 30 dB and 50 dB.

Memory hook: Each ×10 in intensity adds only 10 dB.

Turning a product into a sum

Imagine you must multiply 72 by 12 but you only have a table of logarithms and are told $lo g 2 = 0.3010$ , $lo g 3 = 0.4771$ .

Find $lo g (72 \times 12)$ without multiplying.

Try it first, then show the solutionSolution

$lo g 72 + lo g 12 = 1.8572 + 1.0791$ .
$= 2.9363$ , which is $lo g 864$ (since $72 \times 12 = 864$ ).

Answer: 2.9363

Memory hook: Logs were invented to turn multiplying into adding.

Unknown from log form

Suppose a chemist uses $pH = - lo g [H^{+}]$ and measures a solution with pH 3.

Find the concentration $[H^{+}]$ .

Try it first, then show the solutionSolution

$- lo g [H^{+}] = 3$ , so $lo g [H^{+}] = - 3$ .
$[H^{+}] = 1 0^{- 3} = 0.001$ .

Answer: 10⁻³ (0.001) units.

Memory hook: Log form → exponential form to unlock the number.

Topics in this chapter

Logarithms formula sheet

Definition: $a^{x} = N ⟺ lo g_{a} N = x$; a > 0, a ≠ 1, N > 0.
Special values: $lo g_{a} 1 = 0; lo g_{a} a = 1$; Base 10: log 10 = 1.
Product law: $lo g_{a} mn = lo g_{a} m + lo g_{a} n$; Not log(m + n).
Quotient law: $lo g_{a} \frac{m}{n} = lo g_{a} m - lo g_{a} n$; Not log m ÷ log n.
Power law: $lo g_{a} m^{n} = n lo g_{a} m$; Roots: n = 1/2, 1/3 …
Inverse: $a^{l o g_{a} N} = N$; Exponential undoes log.

How the school exam asks Logarithms

The question patterns to expect, and the method that works for each.

Question type	How it is asked	Method
Interchange forms	Write 2⁻³ = 1/8 in log form, or log₄ 64 = 3 in exponential form.	Base stays the base; the log equals the index.
Evaluate a logarithm	Find log₃ (1/81), log_√2 8, log 0.001.	Let it be x, go to exponential form, equate indices.
Find the unknown	log_x 243 = 5, log₅ x = −2, log₃ (2x + 1) = 2.	Exponential form, then solve.
Expand	Expand log (a⁴b²/c³) or log √(x³y/z²).	Products → sums, quotients → differences, powers → multipliers.
Simplify or combine	Express as a single log, or show that an expression equals log 2.	Multipliers into indices; multiply and divide the numbers.
Given log 2, log 3	Find log 12, log 1.5, log 72, log 0.018.	Prime-factorise; use 5 = 10/2 and powers of 10.
Prove a relation	If a² + b² = 7ab, prove log ((a + b)/3) = ½(log a + log b).	Rearrange the given relation into a product or power, then take logs.

Tricks to remember Logarithms

Short memory aids. Say each one aloud once, then use it on a question the same day.

Log is the index
'log base a of N' = 'which power of a gives N?'
Use it when: Every conversion.
Laws step down
Multiply → add, divide → subtract, power → multiply. Same as the laws of indices.
Use it when: Choosing the law.
Base on the bottom, both ways
In aˣ = N and logₐ N = x, the base a is written low in both.
Use it when: Not mixing up base and number.
Five is ten over two
log 5 = 1 − log 2.
Use it when: Questions that give only log 2 and log 3.

Logarithms: exam-level solved questions

Ordered from exam level to HOTS. Cover the solution, write your own, then compare step by step.

Example 1: Log of a fraction · 1 mark · MCQExam level

The value of $lo g_{3} \frac{1}{81}$ is:
(a) 4
(b) $- 4$
(c) $\frac{1}{4}$
(d) $- \frac{1}{4}$

Try it first, then show the solutionSolution

$\frac{1}{81} = 3^{- 4}$ .

Answer: (b)

Example 2: Unknown base and unknown number · 2 marks · Short answerExam level

Find $x$ : (i) $lo g_{x} 243 = 5$ (ii) $lo g_{3} (2 x + 1) = 2$ .

Try it first, then show the solutionSolution

(i) $x^{5} = 243 = 3^{5}$ , so $x = 3$ .
(ii) $2 x + 1 = 3^{2} = 9$ , so $x = 4$ .

Answer: (i) 3 (ii) 4

Example 3: Using log 2 and log 3 · 3 marks · Short answerExam level

Given $lo g 2 = 0.3010$ and $lo g 3 = 0.4771$ , find (i) $lo g 12$ (ii) $lo g 1.5$ (iii) $lo g 0.18$ .

Try it first, then show the solutionSolution

(i) $2 lo g 2 + lo g 3 = 1.0791$ .
(ii) $lo g 3 - lo g 2 = 0.1761$ .
(iii) $0.18 = \frac{2 \times 9}{100}$ : $lo g 2 + 2 lo g 3 - 2 = 0.3010 + 0.9542 - 2 = - 0.7448$ .

Answer: (i) 1.0791 (ii) 0.1761 (iii) −0.7448

Example 4: Single logarithm · 3 marks · Short answerHarder exam level

Express as a single logarithm: $2 lo g 3 + 3 lo g 2 - lo g 6 - 1$ .

Try it first, then show the solutionSolution

$2 lo g 3 = lo g 9$ , $3 lo g 2 = lo g 8$ , $1 = lo g 10$ .
$lo g \frac{9 \times 8}{6 \times 10} = lo g \frac{72}{60}$ .
$= lo g \frac{6}{5} = lo g 1.2$ .

Answer: log 1.2

Example 5: Solve a log equation · 3 marks · Short answerHarder exam level

Solve $2 lo g x + 1 = lo g 160$ .

Try it first, then show the solutionSolution

$lo g x^{2} + lo g 10 = lo g 160$ .
$lo g 10 x^{2} = lo g 160$ , so $10 x^{2} = 160$ and $x^{2} = 16$ .
$x = 4$ ( $x$ must be positive for $lo g x$ to exist).

Answer: x = 4

Example 6: From a relation to logs · 4 marks · Long answerHarder exam level

If $a^{2} + b^{2} = 7 ab$ (with $a, b > 0$ ), prove that $lo g \frac{a + b}{3} = \frac{1}{2} (lo g a + lo g b)$ .

Try it first, then show the solutionSolution

Add $2 ab$ : $a^{2} + 2 ab + b^{2} = 9 ab$ , so $(a + b)^{2} = 9 ab$ .
$(\frac{a + b}{3})^{2} = ab$ .
Take logs: $2 lo g \frac{a + b}{3} = lo g a + lo g b$ .
Divide by 2: the required result.

Answer: Proved.

Example 7: Three equal ratios · 4 marks · Long answerHOTS / competency

If $\frac{l o g x}{b - c} = \frac{l o g y}{c - a} = \frac{l o g z}{a - b}$ , prove that $x y z = 1$ .

Try it first, then show the solutionSolution

Let each ratio be $k$ : $lo g x = k (b - c)$ , $lo g y = k (c - a)$ , $lo g z = k (a - b)$ .
Add: $lo g x + lo g y + lo g z = k (b - c + c - a + a - b) = 0$ .
$lo g x y z = 0 = lo g 1$ , so $x y z = 1$ .

Answer: Proved.

Case-study practice (Section B style)

Read the situation, then answer every part. In a school exam set on the board pattern this is a 10-mark question split into 3 + 3 + 4 marks.

Checking a science notebook

A student records three results in her notebook using common logarithms (base 10) and the values $lo g 2 = 0.3010$ , $lo g 3 = 0.4771$ . Result 1: $lo g 5 = 0.6990$ . Result 2: $lo g 45 = 1.6532$ . Result 3: she writes ' $lo g (2 + 3) = lo g 2 + lo g 3 = 0.7781$ '.

(i) · 3 marks
Check Result 1 using only $lo g 2$ .
Answer
Answer: Correct
$lo g 5 = lo g 10 - lo g 2 = 1 - 0.3010 = 0.6990$ ✓.
(ii) · 3 marks
Check Result 2 by expanding $lo g 45$ .
Answer
Answer: Correct
$45 = 9 \times 5$ : $2 lo g 3 + lo g 5 = 0.9542 + 0.6990 = 1.6532$ ✓.
(iii) · 4 marks
Explain the error in Result 3 and give the correct statement about $lo g 2 + lo g 3$ .
Answer
Answer: log 2 + log 3 = log 6, not log 5
The product law turns a product into a sum: $lo g 2 + lo g 3 = lo g (2 \times 3) = lo g 6 = 0.7781$ . There is no law for $lo g (2 + 3)$ ; $lo g 5 = 0.6990$ , which differs.

Where the marks are

How a typical answer is split into step marks. Show every step: a right final answer alone rarely earns full marks.

Expand $lo g \frac{a ^{4} \times b ^{2}}{c ^{3}}$ . (3 marks)

Step	Marks
$lo g a^{4} + lo g b^{2} - lo g c^{3}$	1
Apply the power law to each term	1
$4 lo g a + 2 lo g b - 3 lo g c$	1

Evaluate $lo g_{3} 27 + lo g_{2} 16 - lo g_{5} 125 + lo g_{10} 0.01$ . (4 marks)

Step	Marks
$lo g_{3} 27 = 3$	1
$lo g_{2} 16 = 4$	1
$lo g_{5} 125 = 3$ ; $lo g_{10} 0.01 = - 2$	1
$3 + 4 - 3 - 2 = 2$	1

Simplify $lo g \frac{75}{16} - 2 lo g \frac{5}{9} + lo g \frac{32}{243}$ . (5 marks)

Step	Marks
$2 lo g \frac{5}{9} = lo g \frac{25}{81}$	1
Combine: $lo g (\frac{75}{16} \div \frac{25}{81} \times \frac{32}{243})$	1
$= lo g \frac{75 \times 81 \times 32}{16 \times 25 \times 243}$	1
Cancel to get 2	1
Answer: $lo g 2$	1

Where students lose marks in Logarithms

The slips examiners see most often, and what to write instead.

Mistake: Inventing a law for log (m + n)
Students write $lo g (m + n) = lo g m + lo g n$ ; the product law is about $lo g mn$ .
Wrong working
$lo g (2 + 3) = lo g 2 + lo g 3$
Right working
$lo g 2 + lo g 3 = lo g 6$
Do this: Only products, quotients and powers have laws.
Usually costs: The whole question.
Mistake: Treating log m ÷ log n as log m − log n
$\frac{l o g 8}{l o g 2} = 3$ , but $lo g 8 - lo g 2 = lo g 4$ .
Do this: Write numbers as powers of a common base: $\frac{3 l o g 2}{l o g 2} = 3$ .
Usually costs: 1 to 2 marks.
Mistake: Forgetting that a whole number must become a log
In $2 lo g x + 1 = lo g 160$ , students cannot combine '1' with the logs.
Do this: Replace 1 by $lo g 10$ (base 10) before combining.
Usually costs: 1 mark.
Mistake: Not rejecting impossible values
A log of zero or a negative number is undefined, so some solutions of an equation must be rejected.
Do this: Check each answer in the original log expression.
Usually costs: 1 mark.
Mistake: Applying the power law to the wrong part
$lo g 2 x^{3}$ is $lo g 2 + 3 lo g x$ , not $3 lo g 2 x$ .
Do this: The index belongs only to the base it is written on.
Usually costs: 1 mark.

Practice questions on Logarithms

Write each answer on paper first. Open the hint only if you are stuck, and the answer only when you have finished.

Q1 · 1 mark · MCQExam level
If $lo g_{10} x = - 3$ , then $x$ equals:
(a) $- 1000$
(b) $0.001$
(c) $- 0.001$
(d) $1000$
Hint
x = 10⁻³.
Answer
(b)
Q2 · 2 marks · Short answerExam level
Write in exponential form: $lo g_{4} \frac{1}{16} = - 2$ ; write in log form: $(27)^{\frac{2}{3}} = 9$ .
Hint
Base stays the base.
Answer
$4^{- 2} = \frac{1}{16}$ ; $lo g_{27} 9 = \frac{2}{3}$ .
Q3 · 2 marks · Short answerExam level
Evaluate $\frac{l o g 27}{l o g 9}$ and $\frac{l o g 8}{l o g 4}$ .
Hint
Write as powers of 3 and of 2.
Answer
$\frac{3 l o g 3}{2 l o g 3} = \frac{3}{2}$ ; $\frac{3 l o g 2}{2 l o g 2} = \frac{3}{2}$ .
Q4 · 1 mark · Assertion-reasonHarder exam level
Assertion (A): $lo g 20 = 1 + lo g 2$ .
Reason (R): $lo g mn = lo g m + lo g n$ and $lo g 10 = 1$ .
(a) (A) is true and (R) is false.
(b) (A) is false and (R) is true.
(c) Both (A) and (R) are true, and (R) is the correct explanation for (A).
(d) Both (A) and (R) are true, but (R) is not the correct explanation for (A).
Hint
20 = 10 × 2.
Answer
(c)
Q5 · 2 marks · Short answerHarder exam level
Find $x$ if $lo g_{10} x = 2 - lo g_{10} 4$ .
Hint
2 = log 100.
Answer
$lo g x = lo g \frac{100}{4} = lo g 25$ , so $x = 25$ .
Q6 · 3 marks · Short answerHarder exam level
Expand $lo g \frac{3 x ^{2} y ^{4}}{z ^{5}}$ .
Hint
Cube root = power 1/3.
Answer
$\frac{2}{3} lo g x + 4 lo g y - 5 lo g z$ .
Q7 · 3 marks · Short answerExam level
If $lo g x = lo g 3 + 2 lo g 2 - lo g 6$ , find $x$ .
Hint
Combine into one log.
Answer
$lo g \frac{3 \times 4}{6} = lo g 2$ , so $x = 2$ .
Q8 · 3 marks · Short answerHarder exam level
Solve $lo g (x + 1) + lo g (x - 1) = lo g 3$ .
Hint
Product law, then reject invalid roots.
Answer
$(x + 1) (x - 1) = 3 \Rightarrow x^{2} = 4 \Rightarrow x = \pm 2$ . $x = - 2$ makes $lo g (x - 1)$ undefined, so $x = 2$ .
Q9 · 3 marks · Short answerHarder exam level
If $lo g_{10} x = a$ , express $1 0^{2 a - 1}$ in terms of $x$ .
Hint
10ᵃ = x.
Answer
$1 0^{2 a - 1} = \frac{( 1 0 ^{a} ) ^{2}}{10} = \frac{x ^{2}}{10}$ .
Q10 · 4 marks · Long answerHarder exam level
Given $lo g 2 = 0.3010$ and $lo g 3 = 0.4771$ , find $lo g 24$ and $lo g 0.0125$ .
Hint
24 = 2³ × 3; 0.0125 = 1/80.
Answer
$\frac{1}{2} (3 \times 0.3010 + 0.4771) = \frac{1}{2} (1.3801) = 0.69005$ . $lo g 0.0125 = - lo g 80 = - (3 lo g 2 + 1) = - 1.9030$ .
Q11 · 4 marks · Long answerHOTS / competency
If $lo g \frac{x - y}{2} = \frac{1}{2} (lo g x + lo g y)$ , show that $x^{2} + y^{2} = 6 x y$ .
Hint
Remove logs: ((x − y)/2)² = xy.
Answer
$2 lo g \frac{x - y}{2} = lo g x y \Rightarrow \frac{( x - y ) ^{2}}{4} = x y \Rightarrow x^{2} - 2 x y + y^{2} = 4 x y \Rightarrow x^{2} + y^{2} = 6 x y$ .
Q12 · 4 marks · Long answerHOTS / competency
Without tables, show that $lo g_{2} 3$ lies between 1 and 2, and decide which of $lo g_{2} 3$ and $lo g_{3} 5$ is larger by comparing with $\frac{3}{2}$ .
Hint
Compare 3 with 2¹, 2², 2^(3/2); and 5 with 3^(3/2).
Answer
$2^{1} < 3 < 2^{2}$ , so $1 < lo g_{2} 3 < 2$ . $2^{\frac{3}{2}} = 8 \approx 2.83 < 3$ , so $lo g_{2} 3 > \frac{3}{2}$ . $3^{\frac{3}{2}} = 27 \approx 5.20 > 5$ , so $lo g_{3} 5 < \frac{3}{2}$ . Hence $lo g_{2} 3$ is larger.

Test your recall

Answer in your head or on paper first, then open the card. Pulling an answer out of memory makes it stick far better than re-reading.

1.Write aˣ = N in log form.Show answer
logₐ N = x.
2.What is logₐ 1?Show answer
0.
3.What is logₐ a?Show answer
1.
4.State the quotient law.Show answer
logₐ (m/n) = logₐ m − logₐ n.
5.What is the base when none is written?Show answer
10 (common logarithm).
6.Express log 5 using log 2.Show answer
1 − log 2.
7.Evaluate log₂ 64.Show answer
6.
8.Is log (−5) defined?Show answer
No; the number must be positive.

Best order to study it

Practise interchanging exponential and logarithmic forms.
Evaluate logs and find unknown bases or numbers.
Learn and prove the three laws.
Practise expanding and combining expressions.
Use given values of log 2 and log 3; finish with proofs.

Quick revision checklist

The night before the exam: can you do each of these without looking?

Can you convert between aˣ = N and logₐ N = x both ways?
Can you evaluate logs with fractional or surd bases?
Do you know the three laws and that log (m + n) has none?
Can you expand log (a⁴b²/c³) correctly?
Do you write 1 as log 10 when combining?
Can you find log 12, log 1.5, log 5 from log 2 and log 3?
Do you reject values that make a log undefined?

Logarithms: frequently asked questions

Is the change-of-base formula in the Class IX syllabus?

The syllabus lists interchanging forms and the laws of logarithms with expansion. Focus on these; change of base is not named.

Will I need log tables?

The syllabus says certain questions may require mathematical tables, so learn to read them, but most Class IX questions give values such as log 2 = 0.3010.

Why must the base not be 1?

Every power of 1 is 1, so log₁ N would have no single value.

Can a logarithm be negative?

Yes. log 0.01 = −2 because 10⁻² = 0.01. Only the number inside the log must be positive.

How are logarithms linked to indices?

A logarithm is an index. Each law of logarithms comes from a law of indices.

Based on: CISCE ICSE 2028 Mathematics (51) syllabus (Classes IX and X); CISCE Regulations for ICSE Examination Year 2028