Solving Exponential & Log Equations (AA HL)
Most exponential and log equations on IB papers won't come pre-simplified - you first have to spot the structure (a hidden quadratic, a sum of logs, a variable trapped in an index) before any algebra can start. This page walks through the two techniques that unlock nearly every case, with worked examples and the traps that cost the most marks. It's part of the broader Exponentials & Logarithms topic.
18 questions on this sub-topic.
Two moves that solve almost everything
Covered under IB syllabus reference SL2.10: solving equations, both graphically and analytically, including exponential equations, and using technology when no convenient analytic approach exists.
Take logs of both sides
Once the exponential term is completely isolated, apply \(\ln\) (or any convenient base) to both sides to bring the variable down out of the exponent.
Use technology when needed
Some equations, like \(e^x=\sin x\), have no algebraic solution - graph both sides on your GDC and read off the intersection instead.
Need the full syllabus wording, the formula reference table, or GDC-specific steps? See Exponentials & Logarithms.
Worked examples
Solve \(2^{2x} - 5\cdot 2^{x} + 4 = 0.\)
(a)(i) Give the value with \(x<1\).
(a)(ii) Give the value with \(x>1.\)
Worked solution
Note \(2^{2x}=(2^x)^2\); let \(u=2^x\): \(u^2-5u+4=0.\) M1
\((u-1)(u-4)=0\Rightarrow u=1\) or \(u=4.\) A1
\(2^x=1\Rightarrow x=0\); \(2^x=4\Rightarrow x=2.\) A1 A1
Solve \(\ln(x+2) + \ln(x-1) = \ln 4\).
Worked solution
The two terms add, so use the product law \(\ln A+\ln B=\ln(AB)\): \(\ln\big[(x+2)(x-1)\big]=\ln 4.\) M1
(since \(\ln\) is one-to-one): \((x+2)(x-1)=4.\) M1
\(x^2+x-2=4\Rightarrow x^2+x-6=0\Rightarrow (x+3)(x-2)=0.\) A1
Both \(\ln(x+2)\) and \(\ln(x-1)\) require \(x>1\), so reject \(x=-3.\) R1
Therefore \(x=2.\) A1
Common mistakes
- Taking logs before fully isolating the exponential term. You must get the exponential completely alone on one side first - \(\log\) doesn't distribute over addition, so \(\ln(2e^x+3)\) can't be split into separate pieces.
- Ignoring the domain restriction of a logarithm. \(\ln(x-1)\) needs \(x>1\), not \(x>0\) - the restriction applies to whatever sits inside the log, after any shift, and every candidate solution must be checked against it before being accepted.
- Forgetting to convert back to the original variable. After a substitution like \(u=2^x\), solving for \(u\) isn't the end - each value of \(u\) still has to be turned back into a value of \(x\), and it's easy to hand in the \(u\)-values as if they were the answer.
Ready to practise properly?
18 exponential and log equation questions, marked instantly like the real exam.
Quick answers
How do you solve an exponential equation like \(2^{2x} - 5(2^x) + 4 = 0\)?
Substitute \(u = 2^x\) to turn it into a quadratic in \(u\), solve for \(u\), then convert each value of \(u\) back into \(x\) using logs or index laws.
Why do you need to check the domain when solving log equations?
Combining logs can introduce extra solutions that don't satisfy the original equation's domain - the argument of every log must stay positive - so any candidate solution has to be checked against that restriction before it is accepted.