MathsICSE Class 9 · Logarithms

Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

The statement a^x = N is the same as log_a N = x. Logarithms obey log(mn) = log m + log n, log(m/n) = log m − log n and log(mⁿ) = n log m, with log_a 1 = 0 and log_a a = 1. Use them to evaluate, to find unknowns, and to prove relations.

Follow the colours: the base stays the base, the power becomes the answer, and the laws turn multiply, divide and power into add, subtract and multiply.

Illustration drawn for QuestionBuddy.

Understand Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms

Definition. For $a > 0$ , $a \neq = 1$ and $N > 0$ :
$ $a^{x} = N ⟺ lo g_{a} N = x .$ $
A logarithm is an exponent. $5^{3} = 125$ becomes $lo g_{5} 125 = 3$ ; $3^{- 2} = \frac{1}{9}$ becomes $lo g_{3} \frac{1}{9} = - 2$ . If the base is not written, it is 10: $lo g 1000 = 3$ and $lo g 0.001 = - 3$ .

Two special values. $lo g_{a} 1 = 0$ (since $a^{0} = 1$ ) and $lo g_{a} a = 1$ (since $a^{1} = a$ ). Logarithms of zero and negative numbers are not defined.

Laws (same base).
$ $lo g_{a} (mn) = lo g_{a} m + lo g_{a} n, lo g_{a} \frac{m}{n} = lo g_{a} m - lo g_{a} n, lo g_{a} m^{n} = n lo g_{a} m .$ $
Note what they do not say: $lo g (m + n) \neq = lo g m + lo g n$ , and $\frac{l o g m}{l o g n} \neq = lo g m - lo g n$ .

Evaluating. Write the number as a power of the base: $lo g_{2} 32 = 5$ since $32 = 2^{5}$ ; $lo g_{x} 64 = 3$ means $x^{3} = 64$ , so $x = 4$ .

Finding unknowns. Convert to exponential form: $lo g_{5} (2 x - 1) = 2$ gives $2 x - 1 = 25$ , so $x = 13$ . When a log equation involves two logs, combine them first: $lo g x + lo g (x - 3) = 1$ gives $x (x - 3) = 10$ , so $x^{2} - 3 x - 10 = 0$ and $x = 5$ or $x = - 2$ ; reject $x = - 2$ because $lo g (- 2)$ is undefined.

Proving. If $a^{2} + b^{2} = 7 ab$ , then $(a + b)^{2} = a^{2} + b^{2} + 2 ab = 9 ab$ , so $(\frac{a + b}{3})^{2} = ab$ and
$ $lo g \frac{a + b}{3} = \frac{1}{2} (lo g a + lo g b) .$ $

Level. Class IX is a school-exam class. Conversion and evaluation are routine; equations needing a rejected root and the proof above are at the hardest school-exam level.

Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms: formulas to remember

Definition: $a^{x} = N ⟺ lo g_{a} N = x$; Base a > 0, a ≠ 1.
Product law: $lo g_{a} (mn) = lo g_{a} m + lo g_{a} n$; Same base.
Quotient law: $lo g_{a} \frac{m}{n} = lo g_{a} m - lo g_{a} n$; Same base.
Power law: $lo g_{a} m^{n} = n lo g_{a} m$; Exponent comes down.
Special values: $lo g_{a} 1 = 0, lo g_{a} a = 1$; Any valid base.

How to solve Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms questions

Switch between forms, then apply the laws.

1Convert the log statement to exponential form (or the reverse).
2For evaluation, write the number as a power of the base.
3For several logs with the same base, combine with the laws into one log.
4Convert the single log to exponential form and solve.
5Check that every number inside a log is positive; reject roots that fail.
6For proofs, start from the given relation and reach the required log statement.

Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms in real life

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.

A scale built on powers of 10

Suppose a hypothetical scale reports $S = lo g N$ for a quantity $N$ , with $N = 1000$ giving a reading and $N = 2000$ another (taking $lo g 2 = 0.3010$ ).

Find the readings and the increase when the quantity doubles.

Try it first, then show the solutionSolution

$N = 1000$ : $S = lo g 1 0^{3} = 3$ .
$N = 2000$ : $S = lo g 2 + lo g 1000 = 0.3010 + 3 = 3.3010$ .
Doubling adds $lo g 2 = 0.3010$ .

Answer: 3 and 3.3010; each doubling adds 0.3010.

Memory hook: Product becomes sum: multiplying N adds to S.

How many doublings?

Imagine a number that doubles each step, starting at 1.

After how many steps is it 32?

Try it first, then show the solutionSolution

$2^{x} = 32$ , so $x = lo g_{2} 32 = 5$ .

Answer: 5 steps.

Memory hook: A log answers: what exponent?

Tricks to remember Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms

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

Log is an exponent
log_a N = x just means a^x = N.
Use it when: Converting forms.
Times becomes plus
Product to sum, quotient to difference, power to multiple.
Use it when: Using the laws.
Positive inside
Reject any root that makes a log argument ≤ 0.
Use it when: Solving log equations.

Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms: solved examples

Ordered from board level to HOTS. Try each one before you open the solution.

Example 1: Evaluate a log · 1 mark · MCQBoard level

The value of $lo g_{2} 32$ is:
(a) 4
(b) 5
(c) 16
(d) 64

Try it first, then show the solutionSolution

$32 = 2^{5}$ , so $lo g_{2} 32 = 5$ .

Answer: (b) 5

Example 2: Conversion · 3 marks · Short answerBoard level

Write in logarithmic form: (i) $5^{3} = 125$ (ii) $1 0^{- 3} = 0.001$ . Find $x$ if $lo g_{3} x = - 2$ .

Try it first, then show the solutionSolution

(i) $lo g_{5} 125 = 3$ .
(ii) $lo g_{10} 0.001 = - 3$ .
$lo g_{3} x = - 2 \Rightarrow x = 3^{- 2} = \frac{1}{9}$ .

Answer: (i) log₅ 125 = 3 (ii) log₁₀ 0.001 = −3; x = 1/9

Example 3: Unknown inside the log · 3 marks · Short answerHarder board level

Solve (i) $lo g_{5} (2 x - 1) = 2$ (ii) $lo g_{x} 64 = 3$ .

Try it first, then show the solutionSolution

(i) $2 x - 1 = 5^{2} = 25$ , so $x = 13$ .
(ii) $x^{3} = 64$ , so $x = 4$ .

Answer: (i) x = 13 (ii) x = 4

Example 4: Combining logs · 5 marks · Long answerHarder board level

Evaluate $2 lo g 5 + lo g 8 - lo g 2$ and solve $lo g x + lo g (x - 3) = 1$ (base 10).

Try it first, then show the solutionSolution

$2 lo g 5 + lo g 8 - lo g 2 = lo g \frac{25 \times 8}{2} = lo g 100 = 2$ .
$lo g [x (x - 3)] = 1$ , so $x (x - 3) = 10$ .
$x^{2} - 3 x - 10 = 0$ , so $(x - 5) (x + 2) = 0$ .
$x = - 2$ makes $lo g x$ undefined, so $x = 5$ . Check: $lo g 5 + lo g 2 = lo g 10 = 1$ . ✓

Answer: First value 2; x = 5

Example 5: A proof · 5 marks · Long answerHOTS / competency

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

Try it first, then show the solutionSolution

$(a + b)^{2} = a^{2} + b^{2} + 2 ab = 7 ab + 2 ab = 9 ab$ .
So $(\frac{a + b}{3})^{2} = ab$ .
Take logs: $2 lo g \frac{a + b}{3} = lo g a + lo g b$ .
Divide by 2: $lo g \frac{a + b}{3} = \frac{1}{2} (lo g a + lo g b)$ . Hence proved.

Answer: Proved (a, b > 0).

Where students lose marks in Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms

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

Mistake: log(m + n) = log m + log n
The product law is for multiplication.
Wrong working
$lo g (5 + 5) = lo g 5 + lo g 5$ .
Right working
$lo g 10 = 1$ , while $lo g 5 + lo g 5 = lo g 25$ .
Do this: Check which operation is inside.
Usually costs: Whole answer.
Mistake: Forgetting to reject a root
Log arguments must be positive.
Wrong working
$x = 5$ or $x = - 2$ .
Right working
$x = 5$ only.
Do this: Substitute each root into every log.
Usually costs: One mark.
Mistake: Dividing logs for a quotient
log(m/n) is a difference of logs, not a quotient of logs.
Wrong working
$lo g \frac{m}{n} = \frac{l o g m}{l o g n}$ .
Right working
$lo g m - lo g n$ .
Do this: Quotient becomes subtraction.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; unknown-base and unknown-number questions need the conversion to exponential form.
State the law you use at each step.
In an equation with two logs, combine them before converting.
Check that all arguments are positive after solving.

Practice questions on Logarithmic Form, Evaluating Logarithms and the Laws of 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 · MCQBoard level
If $lo g_{2} x = 4$ , then $x$ equals:
(a) 6
(b) 8
(c) 16
(d) 32
Hint
$x = 2^{4}$ .
Answer
(c) 16
Q2 · 3 marks · Short answerHarder board level
Solve $lo g_{3} (x + 2) = 2$ and evaluate $lo g_{5} \frac{1}{25}$ .
Hint
Convert to exponential form.
Answer
$x + 2 = 9$ , $x = 7$ . $lo g_{5} 5^{- 2} = - 2$ .
Q3 · 3 marks · Short answerHarder board level
Expand $lo g \frac{a ^{3} b}{c ^{2}}$ and state the value of the expression $lo g_{10} 1 + lo g_{10} 10$ .
Hint
Powers come down.
Answer
$3 lo g a + \frac{1}{2} lo g b - 2 lo g c$ ; and $0 + 1 = 1$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): $lo g_{a} 1 = 0$ for every valid base $a$ .
Reason (R): $a^{0} = 1$ for every non-zero $a$ .
(a) Both (A) and (R) are true and (R) explains (A).
(b) Both are true but (R) does not explain (A).
(c) (A) is true, (R) is false.
(d) (A) is false, (R) is true.
Hint
Convert log_a 1 = x to a^x = 1.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: A hypothetical scale gives a reading $S = lo g N$ for a quantity $N$ . Take $lo g 2 = 0.3010$ .
(a) Find $S$ when $N = 1000$ .
(b) Find $S$ when $N = 2000$ .
(c) By how much does $S$ rise whenever $N$ doubles?
Hint
$lo g 2000 = lo g 2 + lo g 1000$ .
Answer
(a) 3 (b) 3.3010 (c) 0.3010.

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 a^x = N in log form.Show answer
log_a N = x.
2.What is log_a a?Show answer
1.
3.What is log_a 1?Show answer
0.
4.What does log mⁿ equal?Show answer
n log m.

Logarithmic Form, Evaluating Logarithms and the Laws of Logarithms: frequently asked questions

What is the base when none is written?

In ICSE Class IX it is 10.

Why can the argument not be negative?

No real power of a positive base gives a negative number.

Are tables needed?

Questions give values such as log 2 and log 3, so tables are not needed.