14.6PureStretch

Solving Equations Using Logarithms

When the unknown is stuck in a power, logarithms set it free. Taking logs of both sides and using the power law brings the exponent down where you can solve for it.

30 min Exponentials and Logarithms
Edexcel AS Level Maths: 14.6 Solving Equations Using LogarithmsCurated by Kepler, not produced by us. Watch it before the notes below.
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What you'll be able to do

  • Take logs of both sides of an equation
  • Use the power law to bring down the exponent
  • Solve equations of the form aˣ = b
  • Give answers to a suitable accuracy
1

Take logs of both sides

For an equation like ax=ba^x = b, taking logs of both sides and applying the power law turns the exponent into a multiplier you can divide out.

ax=b  ⇒  xlog⁡a=log⁡b  ⇒  x=log⁡blog⁡aa^{x} = b \;\Rightarrow\; x\log a = \log b \;\Rightarrow\; x = \dfrac{\log b}{\log a}
Logs free the variable from the exponent.
Example 1Solve 3x=203^x = 20 to 3 s.f.
1Take logs: xlog⁡3=log⁡20x\log 3 = \log 20.
2x=log⁡20log⁡3x = \dfrac{\log 20}{\log 3}.
Answerx≈2.73x \approx 2.73

Tip — You can use any base of log (usually log₁₀ or ln) — just be consistent on both sides.

2

Hidden quadratics in eˣ

Some exponential equations are quadratics in disguise — e.g. in e2xe^{2x} and exe^x. Substitute y=exy = e^x, solve the quadratic, then take logs to finish.

Example 1Outline solving e2x−5ex+6=0e^{2x} - 5e^x + 6 = 0.
1Let y=exy = e^x: y2−5y+6=0y^2 - 5y + 6 = 0.
2Factorise: (y−2)(y−3)=0(y - 2)(y - 3) = 0, so ex=2e^x = 2 or ex=3e^x = 3.
3Take logs: x=ln⁡2x = \ln 2 or x=ln⁡3x = \ln 3.
Answerx=ln⁡2, ln⁡3x = \ln 2, \ \ln 3

Formula recap

ax=b⇒x=log⁡blog⁡aa^{x} = b \Rightarrow x = \dfrac{\log b}{\log a}
Solve aˣ = b.
xlog⁡a=log⁡bx\log a = \log b
Power law brings x down.
y=ex substitutiony = e^{x} \text{ substitution}
For hidden quadratics.

Common mistakes to avoid

Writing log b / log a as log(b/a).
They are different! log b ÷ log a is a division of two logs, not log(b/a).
Forgetting the variable can be brought down only via the power law.
Take logs of both sides first, then apply log(aˣ) = x log a.

Key takeaways

  • Take logs of both sides to free a variable from the exponent.
  • aˣ = b ⟹ x = log b / log a.
  • Hidden quadratics in eˣ: substitute y = eˣ, solve, then take logs.

Test yourself

Answer a few questions, then review anything you missed.

Next step

Check your understanding

Work through questions on Solving Equations Using Logarithms. Use the feedback to decide what needs another look.

8 past-paper questions · 19 marks

Practise this topic

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