Solving gives 2^x = 8 or 2^x = −3. What are the solutions?

Exponential equations that become quadratics

Some exponential and logarithmic equations hide a quadratic. A substitution, or changing to one base, brings it out.

International GCSE · 4PM1 · 1B
About 18 minutes

Spot the quadratic

2^2x = (2^x)². So with y = 2^x, the equation 2^2x − 6(2^x) + 8 = 0 becomes
y² − 6y + 8 = 0, so (y − 2)(y − 4) = 0 and y = 2 or 4.
2^x = 2 gives x = 1; 2^x = 4 gives x = 2.
Watch for 2^{x + 1} = 2 × 2^x and 4^x = (2^x)².

Reject impossible values

a^x is always positive. If the quadratic gives 2^x = −3, there is no solution from that root.
Also check the log of a negative number never appears in your answer.

Different bases

log₄ x = log₂ xlog₂ 4 = 12log₂ x.
log₂ x + log₄ x = 6 becomes 32log₂ x = 6, so log₂ x = 4 and x = 16.
For 3^{x + 1} = 5^x, take logs: (x + 1)log 3 = x log 5, so x = log 3(log 5 − log 3) = 2.151.

Worked example

Solve 4^x − 2^{x + 1} − 8 = 0.

Let y = 2^x: 4^x = y² and 2^{x + 1} = 2y.
y² − 2y − 8 = 0, so (y − 4)(y + 2) = 0.
y = −2 is impossible. 2^x = 4, so x = 2.

Answer: x = 2

Key idea

Look for a^2x = (a^x)² and substitute y = a^x. Reject negative values of a^x. Change logs to one base before combining them.

Quick check

Try two quick questions

See if it has sunk in. Answers are marked straight away.

Extra practice

2 more questions

Try each one, then tap to check.

Solve 4^x − 2^{x + 2} − 32 = 0.
Show the answer
Answer: 3
Let y = 2^x. 4^x = y² and 2^{x + 2} = 4y, so y² − 4y − 32 = 0. y = 8 (y must be positive), so x = 3.
Solve log₃ x + log₉ x = 6.
Show the answer
Answer: 81
log₉ x = log₃ x2, so 32 log₃ x = 6. log₃ x = 4 and x = 81.

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Sources and references

Pearson Edexcel International GCSE Further Pure Mathematics (4PM1) specification, topic 1B

Checked by the GCSE-ready editorial team, 17 October 2026. Last updated 2 October 2026. Spotted a mistake? Tell us and we will fix it.