Exponential and Log Equations (AI SL)
These are equations where the unknown sits in the exponent rather than being multiplied by a number - so the usual rearranging tricks don't work. Instead you take logs of both sides, which pulls the exponent down to an ordinary coefficient you can divide by. This page covers the method with worked examples and the mistakes that lose the most marks. It's part of the broader Exponents & Logarithms topic.
28 questions on this sub-topic.
Solving with logs
Covered under IB syllabus reference SL1.5: introducing logarithms with base 10 and \(e\), and the equivalence \(a^x = b \iff \log_a b = x\) for \(a>0,\ b>0\), with \(\log_e x = \ln x\).
What a logarithm means
A logarithm answers "what power gives this value?" AI SL works with base 10 (\(\log\)) and base \(e\) (\(\ln\)) specifically.
\(a^x=b \iff \log_a b = x\), for \(a>0,\ b>0\).
✓ In the formula bookletThe solving technique
Isolate the exponential term first, then take logs of both sides so the log law \(\ln(a^x)=x\ln a\) brings the unknown exponent down as a multiplier.
\(80e^{-0.05t}=40 \Rightarrow t = \dfrac{\ln 0.5}{-0.05} \approx 13.9\).
Not in the formula booklet - techniqueNeed the full syllabus wording and formula-booklet reference table? See Exponents & Logarithms.
Worked examples
Solve \(5^{x} = 100\), giving your answer to 3 significant figures.
Worked solution
\(5^x = 100\). Taking natural logs of both sides and using \(\ln(5^x)=x\ln 5\): \(x\ln 5 = \ln 100.\)M1
\(x = \frac{\ln 100}{\ln 5} = \frac{4.6052}{1.6094} \approx 2.86.\)A1
\(5^2=25,\ 5^3=125\), and 100 sits between, so \(x\) just under 3 is reasonable.
An investment grows as \(V = 1000(1.07)^t.\)
(a) Find the value after 8 years.
(b) Find the time for the investment to double.
Worked solution
(a) Value after 8 years. \(1000(1.07)^8 = 1000(1.71819)\) M1
\(\approx $1718.19.\) A1
(b) Doubling time. \((1.07)^t = 2 \Rightarrow t = \frac{\ln 2}{\ln 1.07} = \frac{0.6931}{0.06766} \approx 10.2 \text{ years}.\)M1 A1 (Rule-of-72 check: \(72/7 \approx 10.3\).)
Common mistakes
- Taking logs before isolating the exponential term. In \(20+60e^{-0.1t}=40\), you must subtract the 20 first to get \(60e^{-0.1t}=20\) before taking logs - taking \(\ln\) of the whole equation as it stands doesn't simplify anything.
- Forgetting the domain of a logarithm. \(\log_a x\) is only defined for \(x>0\) (with \(a>0,\ a\neq1\)) - an answer like \(\ln(-4)\) has no real value and signals an error earlier in the working.
- Rounding too early. Rounding \(\ln 100\) and \(\ln 5\) to 2 decimal places before dividing can shift the final answer outside the accepted range - carry full calculator precision through the division and only round the final answer.
Ready to practise properly?
28 exponential-and-log-equation questions, marked instantly like the real exam. See using your GDC for calculator tips.
Quick answers
How do you solve an equation where the unknown is in the exponent?
Isolate the exponential term on one side, then take logs (base 10 or natural log) of both sides. The log law \(\ln(a^x) = x\ln a\) brings the unknown exponent down as an ordinary multiplier, so you can then divide to solve for it.
What does \(\log_a b = x\) actually mean?
It is just another way of writing \(a^x = b\). The logarithm asks "what power of \(a\) gives \(b\)?" - the answer is \(x\). This equivalence, valid for \(a>0\) and \(b>0\), is how you move between exponential and logarithmic form.