Solving equations and inequalities with x in the index

You can't "undo" a power with ordinary algebra. In $2^x = 5$, dividing by 2 or taking a root doesn't get $x$ on its own.

Logarithms fix this, thanks to the power law:

$\ln(a^u) = u \ln a$

In words: taking $\ln$ brings the power down in front, where it becomes an ordinary multiplier.

1. Get the power on its own on one side, like $a^{\text{something}} = \text{number}$.
2. Take $\ln$ of both sides. (Any log base works. Use $\ln$ or $\lg$, because they're on your calculator.)
3. Bring the power down with $\ln(a^u) = u\ln a$, then solve the ordinary equation that's left.

Quick example: solve $2^x = 5$.

• Take $\ln$ of both sides: $\ln(2^x) = \ln 5$.
• Bring the power down: $x \ln 2 = \ln 5$.
• Divide by $\ln 2$:

$x = \frac{\ln 5}{\ln 2} = 2.32 \text{ (3 s.f.)}$

Check: $2^{2.32} = 4.99\ldots$ ✓

Taking $\ln$ of both sides of an inequality keeps the sign the same way round.

That's because $\ln$ is an increasing function: a bigger input always gives a bigger output. So if $A < B$ then $\ln A < \ln B$. (Both sides must be positive for $\ln$ to exist.)

But watch the next step. If you divide both sides by a negative number, the inequality sign flips.

• $\ln 2$, $\ln 5$ and $\ln 10$ are positive (their inputs are bigger than 1).
• $\ln 0.5$ and $\ln 0.1$ are negative (their inputs are between 0 and 1).

So dividing by $\ln 0.5$ flips the sign. Always test one value at the end to be sure.