Solving an Equation with Different Bases

When an exponential equation has bases that won’t match (here $2$ and $3$), take logs of both sides. The power law $\log(a^k) = k\log a$ brings each exponent down, turning the equation linear in $x$.

Solve $2^x = 3^{x-1}$, giving your answer correct to 3 significant figures. [5]

The working

Step 1, take natural logs of both sides: $\ln(2^x) = \ln(3^{x-1}) \quad \text{(M1)}$

Step 2, apply the power law to bring the exponents down: $x\ln 2 = (x - 1)\ln 3 \quad \text{(M1)}$

Step 3, expand and collect the $x$ terms: $x\ln 2 = x\ln 3 - \ln 3 \implies x\ln 2 - x\ln 3 = -\ln 3$ $x(\ln 2 - \ln 3) = -\ln 3 \quad \text{(M1)}$

Step 4, solve for $x$: $x = \frac{-\ln 3}{\ln 2 - \ln 3} = \frac{\ln 3}{\ln 3 - \ln 2} = \frac{1.0986}{0.4055} \quad \text{(A1)}$ $x = 2.71 \ (3\text{ s.f.}) \quad \text{(A1)}$

Where the marks are won and lost

• Every term gets the log, and the exponent of each side comes down: the right side is $(x-1)\ln 3$, so the bracket must be expanded.
• Collect $x$ terms carefully: $x\ln 2 - x\ln 3 = x(\ln 2 - \ln 3)$. A sign slip here is the usual cause of a wrong answer.
• Any base of log works ($\ln$ or $\lg$); just be consistent on both sides.

Common mistakes

• Forgetting the "$-1$" in the exponent, giving $x\ln 3$ instead of $(x-1)\ln 3$.
• Sign errors when moving $x\ln 3$ across.
• Writing $\frac{\ln 2}{\ln 3}$ or similar, mis-handling the rearrangement.

Topic home: Logarithmic & Exponential Functions pillar. More: Worked examples.