IGCSEFurther MathsEdexcelRevision NotesLogarithms, Indices & SurdsLogarithms, Indices & ExponentialsSolving Exponential Equations

Solving Exponential Equations (Edexcel IGCSE Further Pure Maths): Revision Note

Exam code: 4PM1

Written by: Amber

Reviewed by: Dan Finlay

Updated on 16 October 2024

Solving Exponential Equations

What are exponential equations?

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

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

The change of base law can also be used to solve some exponential equations
- For example,

$\begin{array}{rcl} 27^{x} & = & 9 \end{array}$

Rewrite using the definition of a logarithm

$\begin{array}{rcl} x & = & \log_{27} 9 \end{array}$

Then use the change of base formula from the exam formula sheet
- Use base 3 here because 9 and 27 are both powers of 3

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

In more complicated cases use the laws of logarithms to solve exponential equations

How do I use logarithms to solve exponential equations?

An exponential equation can be solved by taking logarithms of both sides
- $\ln$ , or $\log_{e}$ , is often used
  - Though a log to any base could be used
- The laws of indices may be needed to rewrite the equation first
- The laws of logarithms can then be used to solve the equation
- A question may ask you to give your answer in a particular form
  - For example as an exact value in terms of $\ln$
STEP 1
Take logarithms of both sides

$\begin{array}{rcl} 5^{x} & = & 27 \\ \ln (5^{x}) & = & \ln 27 \end{array}$

STEP 2
Use the laws of logarithms to move powers out of the logarithms

$x \ln 5 = \ln 27$

STEP 3
Rearrange to isolate x

$x = \frac{\ln 27}{\ln 5}$

Note that this is the exact solution to the equation

STEP 4
Use logarithms in your calculator to find the value of $x$

$\begin{array}{rcl} x & = & 2.047818 . . . \\ = & 2.05 (3 s . f .) \end{array}$

Only perform this step if required by the question

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
  - $4^{x} = {(2^{2})}^{x} = 2^{2 x} = {(2^{x})}^{2}$
  - $e^{2 x} = e^{x + x} = e^{x} \times e^{x} = {(e^{x})}^{2}$
- This can be used to factorise quadratic expressions
  - $4^{x} - 2^{x} = {(2^{x})}^{2} - 2^{x} = 2^{x} (2^{x} - 1)$
  - $e^{2 x} - 4 e^{x} - 5 = {(e^{x})}^{2} - 4 e^{x} - 5 = (e^{x} + 1) (e^{x} - 5)$

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: $4^{x} = {(2^{2})}^{x} = 2^{2 x} = {(2^{x})}^{2}$

Also note that $2^{x + 1} = 2^{1} \times 2^{x} = 2 \times 2^{x}$

Substitute these into the equation and simplify

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

Factorise the quadratic

The left-hand side is in the form $y^{2} - 6 y + 9$ where $y = 2^{x}$
This factorises to ${(y - 3)}^{2}$

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

Solve for $2^{x}$

$\begin{array}{rcl} 2^{x} - 3 & = & 0 \\ 2^{x} & = & 3 \end{array}$

Take logarithms of both sides

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

Use $\log_{a} x^{k} = k \log_{a} x$ to take the power out of the logarithm

Remember that $\ln = \log_{e}$

$x \ln 2 = \ln 3$

Solve for $x$

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

Use your calculator to find the decimal equivalent of that exact answer

$x = 1.584962 . . .$

Round to 3 significant figures

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

Once $2^{x} = 3$ is found, the logarithm $x = \log_{2} (3)$ could be used instead to find the value of $x$

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