Solving an Exponential Equation with ln

The move that solves every ”$e$ to the power” equation is: isolate the exponential, then take $\ln$ of both sides so the power drops down. Keep an exact form (with $\ln$) as well as a decimal, because a “solve” question on Paper 2 usually wants both, or specifically the exact one.

Solve $5e^{2x} = 40$, giving your answer in exact form and correct to 3 significant figures. [4]

The working

Step 1, isolate the exponential by dividing by $5$: $e^{2x} = 8 \quad \text{(M1)}$

Step 2, take natural logs of both sides. Since $\ln(e^{2x}) = 2x$: $2x = \ln 8 \quad \text{(M1)}$

Step 3, solve for $x$ (exact form first): $x = \frac{1}{2}\ln 8 \quad \text{(A1, exact)}$

Step 4, evaluate to 3 s.f.: $x = \frac{1}{2}(2.0794\ldots) = 1.04 \quad \text{(A1, 3 s.f.)}$

You may also simplify the exact form: $\ln 8 = \ln 2^3 = 3\ln 2$, so $x = \frac{3}{2}\ln 2$, an equally acceptable exact answer.

Where the marks are won and lost

• Divide before you log. Taking $\ln$ of $5e^{2x} = 40$ directly gives $\ln 5 + 2x = \ln 40$, correct but longer and a common source of slips. Isolate the exponential first.
• The exact form ($\frac12\ln 8$ or $\frac32\ln 2$) is its own mark. A decimal alone can lose it when “exact” is demanded.
• $\ln(e^{2x}) = 2x$, not $2x\ln e \cdot$ anything, remember $\ln e = 1$.

Common mistakes

• Writing $2x = \ln 40 - \ln 5$ but then mishandling it (it does equal $\ln 8$, but only if combined correctly).
• Forgetting the $\frac12$ from the power $2x$.
• Rounding $\ln 8$ too early and losing accuracy in the final figure.

Full method: e^x and ln x notes. See also Solving Log & Exponential Equations. Topic home: Logs & Exponentials pillar.