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Solving Exponential and Logarithmic Equations

Updated July 2026

Solving exponential equations

The solution of equations of the form ax=ba^x = b, and equations which can be reduced to this form, including those that need prior algebraic manipulation; for example, 32x=43^{2x} = 4 and 25x3×5x+2=025^x - 3 \times 5^x + 2 = 0.

Example

Solve: 52x=275^{2x} = 27

We will solve this exactly [exactly means we will find an expression for the value of xx rather than calculating [using a calculator] and then rounding the answer; many log values are irrational [like surds] and so cannot be expressed precisely as a decimal, which is why we often use surds and log expressions rather than rounded numerical values].

We can take logs of both sides, but we need to decide which base is best – we could use base 5 here or [because 27=3327 = 3^3] we could use base 3. We will try both approaches just for completeness.

Approach 1: Take log5\log_5 of both sides:

log552x=log527=log533\log_5 5^{2x} = \log_5 27 = \log_5 3^3

Simplifying:

2x=3log532x = 3 \log_5 3

and so

x=32log53x = \frac{3}{2} \log_5 3

Approach 2: Take log3\log_3 of both sides:

log352x=log327=log333\log_3 5^{2x} = \log_3 27 = \log_3 3^3

Simplifying:

2xlog35=32x \log_3 5 = 3

And so x=32log35x = \frac{3}{2 \log_3 5}.

You can check these two approaches give the same value using the change of base formula.

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