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Exponentials and logarithms · Subtopic 3 of 5

Exponential Equations and Inequalities

Bringing an unknown power down with logarithms: solving ax=ba^x=b, equations with xx in two powers, disguised quadratics, and inequalities like "the smallest nn such that 0.8n<0.10.8^n<0.1".

Cambridge 9709 Paper 2Cambridge 9709 Paper 3Edexcel YMA01 WMA12Edexcel 9MA0 Paper 1Edexcel 9MA0 Paper 2AQA 7357 Paper 1≈ 40 minRevision 6 · updated 2026-10-27
Section 1 of 14

Learning objectives

  • Solve ax=ba^x=b and apx+q=ba^{px+q}=b by taking logarithms.
  • Solve equations with the unknown in two different powers, such as 2x+1=5x2^{x+1}=5^x.
  • Spot and solve disguised quadratics such as 32x−4(3x)+3=03^{2x}-4(3^x)+3=0.
  • Solve exponential inequalities, taking care when dividing by a negative logarithm.
Pause and think

2x=102^x=10. You know 23=82^3=8 and 24=162^4=16. Between which two whole numbers is xx? How could you find it exactly?

Section 2 of 14

Core idea

Start hereWhen will it double?

Money grows at 5%5\% a year: after nn years, 1.05n1.05^n times as much. When does it double? 1.05n=21.05^n=2. The unknown is in the power, where ordinary algebra can't reach. Logarithms bring it down.

The big idea

To solve ax=ba^x=b, take logs of both sides: xln⁡a=ln⁡bx\ln a=\ln b, so x=ln⁡bln⁡ax=\dfrac{\ln b}{\ln a}. Equations in a2xa^{2x} and axa^x are hidden quadratics: substitute u=axu=a^x.

1. Taking logs

3ˣ = 20 where the curve meets the line
−1123510152025300xyy = 3ˣy = 20x ≈ 2.73
32=93^2=9 and 33=273^3=27, so the answer is between 22 and 33, nearer 33. A quick sense-check before using the calculator.
Worked exampleDoubling: 1.05n=21.05^n=2

Solve.

Goal
nn.
Strategy
Logs of both sides; power law.
  1. ln⁡(1.05n)=ln⁡2\ln(1.05^n)=\ln2
    Take ln⁡\ln of both sides (both positive).
  2. nln⁡1.05=ln⁡2n\ln1.05=\ln2
    Power law.
  3. n=ln⁡2ln⁡1.05=14.2n=\frac{\ln2}{\ln1.05}=14.2
    Divide.
AnswerAbout 14.214.2 years (so 15 whole years).
✓ Check1.0514.2=2.001.05^{14.2}=2.00 ✓.
Worked example52x−1=3x+25^{2x-1}=3^{x+2}

Solve to 3 s.f.

Goal
xx.
Strategy
Logs, expand, collect xx terms.
  1. (2x−1)ln⁡5=(x+2)ln⁡3(2x-1)\ln5=(x+2)\ln3
    Take logs; power law.
  2. 2xln⁡5−ln⁡5=xln⁡3+2ln⁡32x\ln5-\ln5=x\ln3+2\ln3
    Expand.
  3. x(2ln⁡5−ln⁡3)=2ln⁡3+ln⁡5x(2\ln5-\ln3)=2\ln3+\ln5
    Collect xx terms on one side.
  4. x=2ln⁡3+ln⁡52ln⁡5−ln⁡3=3.8072.120=1.80x=\frac{2\ln3+\ln5}{2\ln5-\ln3}=\frac{3.807}{2.120}=1.80
    Divide.
Answerx=1.80x=1.80.
✓ CheckWith x=1.796x=1.796: 52.592≈64.85^{2.592}\approx64.8 and 33.796≈64.73^{3.796}\approx64.7 ✓ (equal up to rounding).

2. Hidden quadratics

Worked example22x−5(2x)+4=02^{2x}-5(2^x)+4=0

Solve.

Goal
All xx.
Strategy
22x=(2x)22^{2x}=(2^x)^2; let u=2xu=2^x.
  1. u2−5u+4=0⇒(u−1)(u−4)=0u^2-5u+4=0\Rightarrow(u-1)(u-4)=0
    Substitute; factorise.
  2. 2x=1⇒x=0;2x=4⇒x=22^x=1\Rightarrow x=0;\quad2^x=4\Rightarrow x=2
    Undo.
Answerx=0x=0 or 22.
✓ Checkx=2x=2: 16−20+4=016-20+4=0 ✓.
Worked exampleex−6e−x=1e^x-6e^{-x}=1

Solve.

Goal
xx.
Strategy
Multiply by exe^x to remove the negative power.
  1. e2x−6=exe^{2x}-6=e^x
    e−x⋅ex=1e^{-x}\cdot e^x=1.
  2. u2−u−6=0⇒(u−3)(u+2)=0u^2-u-6=0\Rightarrow(u-3)(u+2)=0
    u=exu=e^x.
  3. ex=3⇒x=ln⁡3;ex=−2 impossiblee^x=3\Rightarrow x=\ln3;\quad e^x=-2\ \text{impossible}
    ex>0e^x>0 always.
Answerx=ln⁡3x=\ln3.
✓ Check3−63=13-\frac63=1 ✓.

3. Inequalities

Worked exampleSmallest nn with 0.8n<0.010.8^n<0.01

Find it.

Goal
Whole nn.
Strategy
Logs; dividing by a negative log flips the sign.
  1. nln⁡0.8<ln⁡0.01n\ln0.8<\ln0.01
    Take logs.
  2. n>ln⁡0.01ln⁡0.8=20.6n>\frac{\ln0.01}{\ln0.8}=20.6
    ln⁡0.8<0\ln0.8<0: flip.
Answern=21n=21.
✓ Check0.820=0.01150.8^{20}=0.0115, 0.821=0.00920.8^{21}=0.0092 ✓.
Common misconceptionForgetting to flip ln⁡\ln of a number between 00 and 11 is negative.
Pause and think

Solve e3x=7e^{3x}=7.

Summary

  • Logs bring powers down.
  • u=axu=a^x for hidden quadratics; reject ax≤0a^x\le0.
  • Flip when dividing by a negative log.
Key words
Exponential equation
Unknown in the exponent.
HistoryDating the past. In 1949 Willard Libby showed that the amount of carbon-14 in dead material falls exponentially, halving about every 57305730 years. Solving an exponential equation then gives the age of bones, wood or cloth. He won the 1960 Nobel Prize in Chemistry for it.
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