Solving Equations with e and ln (AA SL)
Once an equation mixes an unknown with an exponential or a logarithm, isolating that variable means moving between the two forms. This page pulls together the log laws you need to combine or split logarithms, and the routine for taking a natural log to bring an exponent down. It's part of the broader Exponentials & Logarithms topic.
22 questions on this sub-topic.
The tools you need
Covered under IB syllabus reference SL1.7. Both of these come from the formula booklet, so the skill is recognising when to switch forms or combine logs, not memorising them.
Switching forms
\(a^x = b \iff x = \log_a b\)
Use this the moment the unknown is stuck in an exponent - converting to log form (or taking \(\ln\) of both sides) brings it back down to ground level.
Log laws
\(\log_a(xy)=\log_a x+\log_a y,\quad \log_a\!\left(\dfrac{x}{y}\right)=\log_a x-\log_a y,\quad \log_a(x^m)=m\log_a x\)
Use these to combine several logs into one before converting to exponential form, or to pull an exponent out in front where it can be divided away.
Need the full syllabus wording and change-of-base rule? See Exponentials & Logarithms.
Worked examples
Solve \(\log_2(x+6)-\log_2 x=2\).
Worked solution
Quotient law: \(\log_2\dfrac{x+6}{x}=2\). M1
Convert to exponential form: \(\dfrac{x+6}{x}=2^2=4\). A1
Solve: \(x+6=4x\Rightarrow3x=6\Rightarrow x=2\). A1
This is valid since \(x>0\) is needed for the logs to be defined. R1
Let \(f(x)=2e^{x}-3\).
(a) Find \(f^{-1}(x)\).
(b) State its domain.
Worked solution
(a) Set \(y=2e^x-3\) and isolate \(e^x\): \(e^x=\dfrac{y+3}{2}\). M1
Take \(\ln\): \(x=\ln\!\left(\dfrac{y+3}{2}\right)\). A1
So \(f^{-1}(x)=\ln\!\left(\dfrac{x+3}{2}\right)\). A1
(b) The log argument must be positive: \(\dfrac{x+3}{2}>0\Rightarrow x>-3\). M1
\(x>-3\). A1
For \(f(x)=5e^{0.2x}\), find \(x\) when \(f(x)=20\), to 3 significant figures
Worked solution
\(5e^{0.2x}=20\Rightarrow e^{0.2x}=4.\) M1
\(0.2x=\ln4\Rightarrow x=\dfrac{\ln4}{0.2}\) A1 \(\approx6.93\) (3 significant figures). M1A1
Common mistakes
- Not rejecting invalid solutions. Combining logs and solving can produce a root that makes an original log argument negative or zero - always check each solution against the domain of every log in the original equation before accepting it.
- Taking the log of both sides incorrectly. \(\ln(e^x)=x\) is fine, but \(\ln(a+b)\) is not the same as \(\ln a+\ln b\) - only take logs once the exponential term is fully isolated, not partway through an addition.
- Losing the base when converting between forms. \(a^x=b\) becomes \(x=\log_a b\), not \(x=\log b\) - dropping the base \(a\) (or swapping to \(\ln\) without adjusting) gives the wrong numerical answer.
Ready to practise properly?
22 e-and-ln equation questions, marked instantly like the real exam.
Quick answers
How do you solve an equation like log2(x+6) - log2(x) = 2?
Combine the logs into a single one using the quotient law, convert the result to exponential form, then solve the resulting equation for \(x\) - and check the solution keeps every log argument positive.
How do you find the inverse of an equation involving e^x?
Swap \(x\) and \(y\), isolate the \(e^x\) (or \(e^y\)) term on one side, then take the natural log of both sides to bring the exponent down and solve for the remaining variable.