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Exponential Equations (Cambridge (CIE) O Level Additional Maths): Revision Note

Exam code: 4037

Written by: Amber

Reviewed by: Dan Finlay

Updated on 2 July 2026

Solving exponential equations

What are exponential equations?

An exponential equation is an equation where the unknown is a power
- In simple cases the solution can be spotted without the use of a calculator
- For example,

$\begin{array}{rcl} 5^{2 x} & = & 125 \\ 2 x & = & 3 \\ x & = & \frac{3}{2} \end{array}$

In more complicated cases the laws of logarithms should be used to solve exponential equations

The change of base law can be used to solve some exponential equations without a calculator
- For example,

$\begin{array}{rcl} 27^{x} & = & 9 \\ x & = & \log_{27} 9 \\ = & \frac{\log_{3} 9}{\log_{3} 27} \\ = & \frac{2}{3} \end{array}$

How do we use logarithms to solve exponential equations?

An exponential equation can be solved by taking logarithms of both sides
The laws of indices may be needed to rewrite the equation first
The laws of logarithms can then be used to solve the equation
- ln (log_e) is often used
- The answer is often written in terms of ln
A question my ask you to give your answer in a particular form
Follow these steps to solve exponential equations
- STEP 1: Take logarithms of both sides
- STEP 2: Use the laws of logarithms to remove the powers
- STEP 3: Rearrange to isolate x
- STEP 4: Use logarithms to solve for x

What about hidden quadratics?

Look for hidden squared terms that could be changed to form a quadratic
- In particular look out for terms such as
  - 4^x= (2²)^x= 2²^x= (2^x)²
  - e²^x= (e²)^x= (e^x)²

Examiner Tips and Tricks

Always check which form the question asks you to give your answer in, this can help you decide how to solve it
If the question requires an exact value you may need to leave your answer as a logarithm

Worked Example

Solve the equation $4^{x} - 3 (2^{x + 1}) + 9 = 0$ . Give your answer correct to three significant figures.

'Spot' the hidden quadratic by noticing that $4^{x} = {(2^{2})}^{x} = {(2^{x})}^{2}$ .

Rewrite the first term as a power of 2.

${(2^{x})}^{2} - 3 (2^{x + 1}) + 9 = 0$

Rewrite the middle terms using the laws of indices: If $2^{x + 1} = 2^{x} \times 2^{1} = 2 (2^{x})$

$\begin{array}{l} {(2^{x})}^{2} - 3 \times 2 (2^{x}) + 9 = 0 \\ {(2^{x})}^{2} - 6 (2^{x}) + 9 = 0 \end{array}$

USing a substitution can make this easier to solve.

Let $u = 2^{x}$

${(u)}^{2} - 6 (u) + 9 = 0$

Factorise.

$\begin{array}{l} u^{2} - 6 u + 9 = 0 \\ (u - 3) (u - 3) = 0 \end{array}$

Solve to find u and substitute 2^xback in.

$\begin{array}{rcl} u & = & 3 \\ 2^{x} & = & 3 \end{array}$

Solve the exponential equation 2^x= 3 by taking logarithms of both sides.

$\ln 2^{x} = \ln 3$

Bring the power down using the law of logs $\ln x^{m} = m \ln x$ .

$x \ln 2 = \ln 3$

Rearrange and solve.

$x = \frac{\ln 3}{\ln 2} = 1.584 . . .$

$x = 1.58 (3 s . f .)$

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