MathsICSE Class 9 · Indices (Exponents)

Solving Exponential Equations (ICSE Class 9)

Updated 17 October 2026 · Fact-checked

To solve an exponential equation, write both sides with the same base and equate the exponents: if aˣ = aʸ (a > 0, a ≠ 1) then x = y. When a sum or difference of powers appears, factor out the smallest power or substitute y = aˣ to get a quadratic.

Rewrite both sides with the same base, then equate the exponents to get a simple linear equation.

Illustration drawn for QuestionBuddy.

Understand Solving Exponential Equations

Key principle. If $a^{x} = a^{y}$ with $a > 0$ and $a \neq = 1$ , then $x = y$ . So the aim is always to make both sides a power of the same base.

Steps. (1) Write each side with a prime base. (2) Use the laws to reduce each side to a single power. (3) Equate exponents and solve the resulting linear equation. (4) Check in the original equation.

*Example.* $4^{x + 1} = 8^{x - 1}$ gives $2^{2 x + 2} = 2^{3 x - 3}$ , so $2 x + 2 = 3 x - 3$ and $x = 5$ . Check: $4^{6} = 4096 = 8^{4}$ .

Constants. $a^{x} = 1$ gives $x = 0$ ; $5^{x} = \frac{1}{125}$ gives $5^{x} = 5^{- 3}$ and $x = - 3$ .

Sums of powers. If the equation has addition or subtraction, factor out the smallest power: $2^{x + 2} - 2^{x} = 24$ becomes $2^{x} (4 - 1) = 24$ , so $2^{x} = 8$ and $x = 3$ .

Quadratic in disguise. For $9^{x} - 4 \cdot 3^{x} + 3 = 0$ write $9^{x} = (3^{x})^{2}$ and set $y = 3^{x}$ : $y^{2} - 4 y + 3 = 0$ , so $y = 1$ or $y = 3$ , giving $x = 0$ or $x = 1$ . Remember $y = 3^{x}$ is always positive, so reject non-positive $y$ .

Mixed bases. Collect like powers: for $(\frac{2}{3})^{x} (\frac{3}{2})^{2 x} = \frac{81}{16}$ , write $(\frac{2}{3})^{x} = (\frac{3}{2})^{- x}$ , so the left side is $(\frac{3}{2})^{x}$ . Since $\frac{81}{16} = (\frac{3}{2})^{4}$ , we get $x = 4$ .

Halving and doubling models. A quantity that halves every $h$ hours is $A = A_{0} (\frac{1}{2})^{t / h}$ . To find $t$ set $\frac{A _{0}}{A} = 2^{t / h}$ and equate exponents.

Level. Class IX is a school-exam class. Equal-base equations are routine; the quadratic-in-disguise and model items are at the hardest school-exam level.

Solving Exponential Equations: formulas to remember

Equal bases: $a^{x} = a^{y} \Rightarrow x = y (a > 0, a \neq = 1)$; The core principle.
Constant 1: $a^{x} = 1 \Rightarrow x = 0$; a ≠ 1.
Factor the smallest power: $2^{x + 2} - 2^{x} = 2^{x} (4 - 1)$; Sums of powers.
Substitution: $9^{x} = (3^{x})^{2}$; Let y = 3ˣ, y > 0.
Halving model: $A = A_{0} (\frac{1}{2})^{t / h}$; Half-life style.

How to solve Solving Exponential Equations questions

Same base, equate exponents, solve, check.

1Write all numbers as powers of a prime.
2Reduce each side to one power using the laws.
3If sums appear, factor out the smallest power or substitute.
4Equate exponents and solve the linear (or quadratic) equation.
5Reject any solution that makes a base-power equal a non-positive number.
6Substitute back to check.

Solving Exponential Equations 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.

Medicine in the body

Suppose a dose of 640 mg of a medicine halves in the body every 4 hours (hypothetical model), so the amount after $t$ hours is $640 \times (\frac{1}{2})^{t /4}$ .

When will 5 mg remain?

Try it first, then show the solutionSolution

$\frac{640}{5} = 128 = 2^{7}$ .
$2^{t /4} = 2^{7}$ , so $t = 28$ hours.

Answer: 28 hours.

Memory hook: Divide, write as a power of 2, equate exponents.

Doubling balance

Imagine a savings balance that doubles each period, and a goal of 8 times the starting sum after $t$ periods.

Find $t$ .

Try it first, then show the solutionSolution

$2^{t} = 8 = 2^{3}$ , so $t = 3$ .

Answer: 3 periods.

Memory hook: Write the target as a power of the base.

Tricks to remember Solving Exponential Equations

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

Same base first
You can only equate exponents when the bases match.
Use it when: Every exponential equation.
Factor, do not split
In a sum of powers, factor out the smallest one.
Use it when: 2^(x+2) − 2^x type.
y is positive
In substitutions like y = 3ˣ, y cannot be zero or negative.
Use it when: Rejecting roots.

Solving Exponential Equations: solved examples

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

Example 1: Simple equation · 1 mark · MCQBoard level

If $2^{x} = 32$ , then $x$ equals:
(a) 4
(b) 5
(c) 6
(d) 16

Try it first, then show the solutionSolution

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

Answer: (b) 5

Example 2: Two equations · 3 marks · Short answerBoard level

Solve (i) $3^{2 x - 1} = 27$ (ii) $5^{x} = \frac{1}{125}$ .

Try it first, then show the solutionSolution

(i) $3^{2 x - 1} = 3^{3}$ , so $2 x - 1 = 3$ and $x = 2$ .
(ii) $5^{x} = 5^{- 3}$ , so $x = - 3$ .

Answer: (i) x = 2 (ii) x = −3

Example 3: Different bases · 3 marks · Short answerHarder board level

Solve $4^{x + 1} = 8^{x - 1}$ and verify.

Try it first, then show the solutionSolution

$2^{2 (x + 1)} = 2^{3 (x - 1)}$ , so $2 x + 2 = 3 x - 3$ .
$x = 5$ .
Check: $4^{6} = 4096$ and $8^{4} = 4096$ . ✓

Answer: x = 5

Example 4: A sum of powers · 4 marks · Short answerHarder board level

Solve $2^{x + 2} - 2^{x} = 24$ .

Try it first, then show the solutionSolution

$2^{x} (2^{2} - 1) = 24$ , so $2^{x} \cdot 3 = 24$ .
$2^{x} = 8 = 2^{3}$ , so $x = 3$ .
Check: $2^{5} - 2^{3} = 32 - 8 = 24$ . ✓

Answer: x = 3

Example 5: Quadratic in 3^x · 5 marks · Long answerHOTS / competency

Solve $9^{x} - 4 \cdot 3^{x} + 3 = 0$ .

Try it first, then show the solutionSolution

Let $y = 3^{x}$ (so $y > 0$ ). Then $9^{x} = y^{2}$ .
$y^{2} - 4 y + 3 = 0$ , so $(y - 1) (y - 3) = 0$ and $y = 1$ or $y = 3$ .
$3^{x} = 1$ gives $x = 0$ ; $3^{x} = 3$ gives $x = 1$ .
Check $x = 0$ : $1 - 4 + 3 = 0$ . $x = 1$ : $9 - 12 + 3 = 0$ . ✓

Answer: x = 0 or x = 1

Where students lose marks in Solving Exponential Equations

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

Mistake: Equating when bases differ
You cannot compare exponents of different bases.
Wrong working
$4^{x + 1} = 8^{x - 1} \Rightarrow x + 1 = x - 1$ .
Right working
$2^{2 x + 2} = 2^{3 x - 3}$ , so $x = 5$ .
Do this: Convert to a common prime base.
Usually costs: Whole answer.
Mistake: Losing a root in a quadratic
Both values of y may be valid.
Wrong working
$y = 3$ only, so $x = 1$ .
Right working
$y = 1$ and $y = 3$ , so $x = 0$ and $x = 1$ .
Do this: Solve fully, reject only y ≤ 0.
Usually costs: One mark.
Mistake: Wrong factoring of a sum
2^(x+2) − 2^x is not 2^2.
Wrong working
$2^{x + 2} - 2^{x} = 2^{2}$ .
Right working
$2^{x} (4 - 1)$ .
Do this: Factor out the smallest power.
Usually costs: Whole answer.

Exam tips

Class IX is examined in school papers; substitution questions are the hardest in this chapter.
Always verify by substituting; exponents have no extraneous roots here, but arithmetic slips are common.
Write "equating exponents" as a step in your answer.
For model questions, show the division that gives a power of 2.

Practice questions on Solving Exponential Equations

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 $7^{x} = 1$ , then $x$ is:
(a) 0
(b) 1
(c) 7
(d) $- 1$
Hint
$7^{0} = 1$ .
Answer
(a) 0
Q2 · 3 marks · Short answerHarder board level
Solve $2 7^{x} = 9^{x + 1}$ .
Hint
Write as powers of 3.
Answer
$3^{3 x} = 3^{2 x + 2}$ , so $3 x = 2 x + 2$ and $x = 2$ (check $2 7^{2} = 729 = 9^{3}$ ).
Q3 · 4 marks · Short answerHarder board level
Solve $(\frac{2}{3})^{x} (\frac{3}{2})^{2 x} = \frac{81}{16}$ .
Hint
$(\frac{2}{3})^{x} = (\frac{3}{2})^{- x}$ .
Answer
LHS $= (\frac{3}{2})^{2 x - x} = (\frac{3}{2})^{x}$ . $\frac{81}{16} = (\frac{3}{2})^{4}$ , so $x = 4$ .
Q4 · 1 mark · Assertion-reasonHarder board level
Assertion (A): If $5^{x} = 5^{y}$ then $x = y$ .
Reason (R): For a base greater than 1, a power takes each positive value for exactly one exponent.
(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
One-to-one property.
Answer
(a)
Q5 · 4 marks · Case-basedHOTS / competency
Case study: The amount of a medicine in the body is $640 \times (\frac{1}{2})^{t /4}$ mg after $t$ hours (hypothetical model).
(a) How much remains after 8 hours?
(b) When does 20 mg remain?
(c) When does 5 mg remain?
Hint
Divide 640 by the amount to get a power of 2.
Answer
(a) $640 \times \frac{1}{4} = 160$ mg (b) $\frac{640}{20} = 32 = 2^{5}$ , so $t = 20$ hours (c) $\frac{640}{5} = 128 = 2^{7}$ , so $t = 28$ hours.

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.If aˣ = aʸ, what follows?Show answer
x = y (a > 0, a ≠ 1).
2.Solve aˣ = 1 for a ≠ 1.Show answer
x = 0.
3.What is 9ˣ in terms of 3ˣ?Show answer
(3ˣ)².
4.Why reject y = 0 in y = 3ˣ?Show answer
A positive base never gives 0.

Solving Exponential Equations: frequently asked questions

Is the substitution method in the ICSE Class IX syllabus?

The chapter lists exponential equations; substitution is a natural extension and is given here as an advanced item.

What if the bases are 4 and 8?

Write both as powers of 2.

Can an exponential equation have no solution?

Yes, for example 2ˣ = −4 has no real solution.