Exponential Equations | CIE IGCSE Additional Maths Revision Notes 2025

Solving 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,

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

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

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= 2^2x= (2^x)²
  - e^2x= (e²)^x= (e^x)²

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

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 $2^{x + 1} = 2^{x} \times 2^{1} = 2 (2^{x})$

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

USing a substitution can make this easier to solve.

Let $u = 2^{x}$

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

Factorise.

$u^{2} - 6 u + 9 = 0 (u - 3) (u - 3) = 0$

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.

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

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