Exponential & Log Equations (AA SL)
This is where index laws and log laws come together to actually solve for \(x\). Some equations match to a common base and drop straight to a linear equation; others need the logs combined first and the domain checked afterwards. Knowing which route a question wants is the real skill. It's part of the broader Logs & Exponents topic.
35 questions on this sub-topic.
Two solving methods
Covered under IB syllabus reference SL1.7: laws of exponents with rational exponents, the laws of logarithms \(\log_a(xy)=\log_a x+\log_a y\), \(\log_a(\tfrac{x}{y})=\log_a x-\log_a y\), \(\log_a(x^m)=m\log_a x\), change of base, and solving exponential equations including using logarithms.
Common-base method
If \(a^{f(x)} = a^{g(x)}\) then \(f(x) = g(x)\).
Rewrite both sides as powers of the same base first (e.g. \(8=2^3\)), then equate the exponents and solve the resulting equation.
Log method
Combine logs to \(\log_a A = c \iff A = a^c\), or take \(\log\) of both sides when a common base isn't possible.
Always finish by checking every logarithm's argument is positive at your solution - reject any root that fails this.
Need the individual log laws first, or index laws? See Laws of Logarithms or Index Laws.
Worked examples
Solve \(\log_2 x + \log_2(x-2) = 3\).
Worked solution
Product law \(\log_b M+\log_b N=\log_b(MN)\): \(\log_2\big(x(x-2)\big)=3.\) M1
\(\log_2 A=3\Leftrightarrow A=2^3\), so \(x(x-2)=8\Rightarrow x^2-2x-8=0.\) M1 A1
\((x-4)(x+2)=0\Rightarrow x=4\) or \(x=-2.\) A1
Logs need positive arguments, so reject \(x=-2\). Therefore \(x=4.\) R1
Solve \(2^{x+1} = 8^{x-1}\).
Worked solution
\(8=2^3\), so \(8^{x-1}=2^{3(x-1)}.\) M1
(same base): \(x+1=3(x-1)=3x-3.\) A1 R1
\(4=2x\Rightarrow x=2.\) A1
Solve the system \(\log_2 x + \log_2 y = 5\) and \(\log_2 x - \log_2 y = 1.\)
Worked solution
\((\log_2 x+\log_2 y)+(\log_2 x-\log_2 y)=5+1\Rightarrow 2\log_2 x\) M1 \(=6.\) A1
\(\log_2 x=3\Rightarrow x=8.\) A1
\(2\log_2 y=4\Rightarrow \log_2 y=2\Rightarrow y\) M1 \(=4.\) A1 So \((x,y)=(8,4).\) 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 to reject invalid roots. After solving a log equation, always check each solution makes every logarithm's argument positive - a root like \(x=-2\) in \(\log_2(x-2)\) must be discarded.
- Equating exponents without a shared base. \(a^{f(x)}=b^{g(x)}\) can only be turned into \(f(x)=g(x)\) once both sides are rewritten with the same base - trying to equate the exponents beforehand just produces a wrong equation.
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35 exponential and log equation questions, marked instantly like the real exam.
Quick answers
How do I solve an exponential equation like \(2^{3x-1} = 32\)?
Rewrite both sides with the same base (\(32 = 2^5\)), then equate the exponents: \(3x-1 = 5\), giving \(x = 2\). If a common base isn't possible, take logarithms of both sides instead.
Why do I sometimes have to reject a solution to a log equation?
A logarithm is only defined when its argument is positive. Squaring or combining logs during solving can introduce extra algebraic roots that make an original argument negative or zero - those roots must be discarded.