Using Logs to Solve an Exponential Equation

When an exponential has a base that isn’t $e$, take logs of both sides and use the power law ($\log a^x = x\log a$) to bring the exponent down. Dividing the two logs is effectively the change-of-base formula, and $\ln$ or $\lg$ both work.

Solve $2^x = 7$, giving your answer correct to 3 significant figures. [3]

The working

Step 1, take logs of both sides (using $\ln$, though $\lg$ works equally): $\ln(2^x) = \ln 7 \quad \text{(M1)}$

Step 2, apply the power law to bring the exponent down: $x \ln 2 = \ln 7 \quad \text{(M1)}$

Step 3, solve for $x$: $x = \frac{\ln 7}{\ln 2} = \frac{1.9459\ldots}{0.6931\ldots} = 2.81 \ (3 \text{ s.f.}) \quad \text{(A1)}$

Where the marks are won and lost

• The power law is the key: $\ln(2^x) = x\ln 2$, which frees $x$ from the exponent. Without it you can’t isolate $x$.
• $x = \frac{\ln 7}{\ln 2}$ is a division of logs, not $\ln 7 - \ln 2$ (that would be $\ln\frac72$, a different value). This is the classic error.
• $\ln$ and $\lg$ give the same $x$ because the base cancels in the ratio, use whichever is convenient.

Common mistakes

• Writing $x = \ln 7 - \ln 2$ (confusing the change-of-base ratio with the subtraction law).
• Forgetting the power law and leaving $x$ trapped in the exponent.
• Rounding the two logs too early before dividing.

Full method: Solving Log & Exponential Equations notes. Topic home: Logs & Exponentials pillar.