Exponential and Log Equations (AA HL)
Solving an exponential or logarithmic equation is about turning an unfamiliar-looking equation into one you already know how to handle - usually a linear or quadratic equation in disguise. This page walks through the main techniques, two full worked examples, and the mistakes that cost marks even when the method is right. It's part of the broader Exponents & Logarithms topic.
40 questions on this sub-topic.
Solving techniques
Covered under IB syllabus reference SL1.7: laws of exponents with rational exponents, laws of logarithms, change of base, and solving exponential equations using logarithms.
Same-base exponential equations
If \(a^{f(x)} = a^{g(x)}\), then \(f(x) = g(x)\).
Rewrite both sides as powers of the same base first (e.g. \(27 = 3^3\)), then equate the exponents and solve. This isn't a booklet formula - it follows directly from exponents being one-to-one.
Different-base exponential equations
\(a^x = b \iff x = \dfrac{\ln b}{\ln a}\)
When the bases can't be matched, take logs of both sides. The change-of-base result behind this is in the formula booklet under logarithms.
Need the full syllabus wording and formula-booklet reference table? See Exponents & Logarithms. For calculator strategy on these equations, see the parent topic's GDC guidance.
Worked examples
Giving exact answers:
(a) Solve \(5^{2x-1} = 5^{x+3}.\)
(b) Solve \(3^{x} = 27^{x-2}.\)
Worked solution
First equation:
equal bases \(\Rightarrow 2x - 1 = x + 3\) M1
\(x = 4.\) A1
Second equation:
\(27 = 3^3\), so \(3^x = 3^{3(x-2)} \Rightarrow x = 3x - 6\) M1
\(x = 3.\) A1
Solve \(\log_3(x+5) + \log_3(x-3) = 2.\)
Worked solution
\(\log_3[(x+5)(x-3)] = 2 \Rightarrow (x+5)(x-3)\) M1 \(= 9.\) A1
\(x^2 + 2x - 24 = 0\) M1 \((x+6)(x-4) = 0.\) A1 Reject \(x = -6\) (need \(x > 3\)); so \(x = 4.\) A1
Solve \(3^{2x-1}=27.\)
Worked solution
\(27 = 3^3\) M1
\(\Rightarrow 2x - 1 = 3\) A1
\(\Rightarrow x = 2.\) A1
Common mistakes
- Splitting \(\log(x+y)\) into \(\log x + \log y\). Logarithms don't distribute over addition or subtraction - the product/quotient laws only apply to multiplication and division inside the log.
- Forgetting the domain of a logarithm. \(\log_a x\) is only defined for \(x>0\) - always check each solution against the original equation and reject any that make an argument zero or negative.
- Writing \(2^{2x}\) as \(2 \cdot 2^x\) instead of \((2^x)^2\). This mistake blocks the substitution \(y=2^x\) that turns equations like \(2^{2x}-5(2^x)+4=0\) into a solvable quadratic in \(y\).
Ready to practise properly?
40 exponential and log equation questions, marked instantly like the real exam.
Quick answers
How do you solve an exponential equation like \(5^{2x-1}=5^{x+3}\)?
Once both sides are written as powers of the same base, the exponents must be equal, so you can drop the base and solve the resulting equation in \(x\).
Why do you need to check solutions when solving log equations?
A logarithm is only defined for a positive argument, so any solution that makes the original expression inside a log negative or zero must be rejected, even if it satisfies the algebra.