Solving Exponential/Log Equations (AI HL)
Once a growth or decay model is set up, the real work is solving for the unknown exponent - which almost always means taking logs. This page focuses on that mechanical step: isolating the exponential term, choosing a sensible log, and spotting when an equation is secretly a quadratic. It's part of the broader Exponential & Logarithmic Models topic.
41 questions on this sub-topic.
Taking logs to solve
Covered under IB syllabus reference AHL2.9, which asks you to build and use exponential and natural log models. Solving them for \(x\) relies on two moves.
Isolate, then take logs
\(a^x=b \Rightarrow x=\dfrac{\ln b}{\ln a}\)
Get the exponential term alone on one side first, then apply \(\ln\) (or \(\log\)) to both sides and use the power law to bring the exponent down.
Spot the disguised quadratic
\(e^{2x}-5e^x+6=0 \Rightarrow u=e^x\)
If the equation has an \(e^{2x}\) term and an \(e^x\) term, substitute \(u=e^x\), solve the resulting quadratic in \(u\), then convert each positive root back with \(\ln\).
Need the surrounding model-building and half-life work? See Exponential & Logarithmic Models, including its GDC guidance for solving equations graphically.
Worked examples
Solve \(3^{2x-1} = 20\), giving the answer to 3 significant figures.
Worked solution
Take logs: \((2x-1)\ln 3 = \ln 20\) M1
\(\Rightarrow 2x - 1 = \dfrac{\ln 20}{\ln 3} \approx 2.727.\) A1
\(2x = 3.727\) M1
\(\Rightarrow x \approx 1.86.\) A1
Solve \(e^{2x} - 5e^{x} + 6 = 0\).
(a)(i) Give the value with \(x<0.9\).
(a)(ii) Give the value with \(x>0.9\).
Worked solution
Let \(u = e^x\): \(u^2 - 5u + 6 = 0.\) M1
\((u-2)(u-3) = 0 \Rightarrow u = 2\) or \(3.\) A1
\(e^x = 2 \Rightarrow x = \ln 2;\) A1 \(e^x = 3 \Rightarrow x = \ln 3.\) A1
Common mistakes
- Taking logs before isolating the exponential. \(\ln\) can only simplify a power once the exponential term stands alone on one side - taking logs of \(2e^x+3=11\) directly does nothing useful.
- Missing the second root of a disguised quadratic. After substituting \(u=e^x\), a quadratic in \(u\) usually has two solutions; both must be converted back to \(x\) unless one gives \(u \le 0\) (which has no real solution, since \(e^x>0\) always).
- Forgetting the domain of a log model. A natural log model \(f(x)=a+b\ln x\) is undefined for \(x\le0\) - always state \(x>0\) when asked for the domain.
Ready to practise properly?
40 exponential/log-equation questions, marked instantly like the real exam.
Quick answers
How do you solve an exponential equation like a^x = b?
Take logs (usually natural log) of both sides: \(\ln(a^x) = \ln b\), so \(x \ln a = \ln b\), giving \(x = \dfrac{\ln b}{\ln a}\).
What do I do if an exponential equation looks like a quadratic?
Substitute \(u = e^x\) (or \(u=a^x\)) to turn it into an ordinary quadratic in \(u\), solve for \(u\), then convert each positive value of \(u\) back to \(x\) using logs.