Solving Exponential Equations (AI SL)
Once the unknown sits in the exponent, ordinary algebra can't reach it - you need a log to bring it back down. This page focuses on that one move: isolating the exponential term, then taking a log of both sides to solve for \(x\) or \(t\). It's part of the broader Exponential Models topic.
11 questions on this sub-topic.
The key move
Covered under IB syllabus reference SL2.5, which describes the exponential growth and decay models \(f(x)=ka^x+c\) and \(f(x)=ke^{rx}+c\) used in contexts like population growth, radioactive decay and compound interest. Solving them for the unknown exponent always comes back to the same log identity.
General base
\(a^x=b \iff x=\log_a b\)
Isolate the power first, then convert to a logarithm. This is the version your GDC's log key handles directly.
Base \(e\)
\(e^{kx}=b \iff kx=\ln b\)
The same rule with \(a=e\) - use natural log (ln) rather than \(\log_{10}\) whenever the model is written with \(e\).
Need the full syllabus wording and formula-booklet reference table? See Exponential Models.
Worked examples
Bacteria follow \(N(t)=200(2)^{t/3}\), \(t\) in hours.
(a) Find the number after 6 hours.
(b) Find when the count reaches \(3200.\)
Worked solution
(a) Number after 6 hours. \(6/3=2\) doubling periods: \(N(6)=200(2)^{6/3}=200(2)^2=200(4)=800.\) A1
(b) When the count reaches \(3200\). \(200(2)^{t/3}=3200\Rightarrow2^{t/3}=16=2^4\Rightarrow t/3=4\Rightarrow t=12\) hours. A1
Town A: \(P_A=5000(1.02)^t.\) Town B: \(P_B=3000(1.05)^t.\)
(a) Find each population at \(t=10.\)
(b) Find when the two populations are equal (to 3 significant figures).
Worked solution
(a) Populations at \(t=10\). \(P_A(10)=5000(1.02)^{10}\approx6095,\qquad P_B(10)=3000(1.05)^{10}\approx4887.\) A1
(b) When equal. Set \(P_A=P_B\): \(5000(1.02)^t=3000(1.05)^t\Rightarrow\dfrac{5}{3}=\left(\dfrac{1.05}{1.02}\right)^t\). M1
Take logs: \(t=\frac{\ln(5/3)}{\ln(1.05/1.02)}\approx17.6\text{ years}.\) A1
A shop's stock level is modelled by \(S(x)=5(3)^{x}\) items, where \(x\) is the number of weeks since the model started. Find the value of \(x\) for which \(S(x)=135.\)
Worked solution
\(5(3)^x=135\Rightarrow(3)^x=27\): M1
\(3^x=3^3\Rightarrow x=3\) weeks. A1
Common mistakes
- Sign errors when isolating the exponential term before taking logs. Rearranging \(k+c-\text{value}\) incorrectly is a common slip that produces a negative or undefined logarithm.
- Taking \(\log_{10}\) of an \(e\)-based model instead of \(\ln\). Both work algebraically, but mixing them on a calculator mid-solution is where arithmetic slips creep in - pick one log and stick with it throughout.
- Rounding the exponent too early. Rounding \(t\) to 1 or 2 significant figures before substituting back can shift the final answer outside the accepted tolerance - carry extra decimal places until the last line.
Ready to practise properly?
11 exponential-equation questions, marked instantly like the real exam.
Quick answers
How do you solve an exponential equation like \(a^x = b\)?
Isolate the exponential term on one side, then take logs of both sides: \(a^x=b\) is equivalent to \(x=\log_a b\). On a GDC you can also solve it directly, or plot both sides and read off the intersection - see the parent topic's GDC guidance.
Why do you take natural logs when the base is \(e\)?
Taking \(\ln\) of both sides cancels the base-\(e\) exponential directly, since \(\ln(e^{kx})=kx\). It's the same log rule as \(a^x=b\), just with \(a=e\).