Solving Equations Using Logarithms
Four recurring situations, one underlying strategy: use a logarithm (or an exponential) to free the unknown, then always check the answer is actually valid.
Four situations, one strategy
Every equation involving logs or exponentials in this course reduces to one of four patterns: the unknown sits alone in an exponent ($a^x = b$), the unknown appears twice in a way that hides a quadratic ($e^{2x} - 5e^x + 6 = 0$), several logs need combining before they can be removed, or the same unknown appears on both sides with different bases. In every case the strategy is the same: use logs (or $e$) to strip away whichever operation is trapping the unknown, then check the result actually makes sense — a solution that makes a logarithm's argument zero or negative must be rejected, however cleanly it fell out of the algebra.
Type 1 — a^x = b
Take logs of both sides, then use the power law to bring the exponent down to a multiplier:
Any base of log works, as long as you use the same base on both sides — the answer comes out identical either way.
Worked example 1
When the exponent is a bracket like $2x-1$ rather than just $x$, the whole bracket comes down together — don't split $2x-1$ apart before taking logs. Once you have $2x-1 = (\text{number})$, finish it as an ordinary linear equation.
Type 2 — a quadratic in aˣ or eˣ
If the unknown appears as both $e^x$ and $e^{2x}$ (or $a^x$ and $a^{2x}$), the equation is a quadratic in disguise, because $e^{2x} = (e^x)^2$. Substitute $y = e^x$, solve the ordinary quadratic in $y$, then undo the substitution with $\ln$.
$e^x$ (and $a^x$ for $a>0$) is always positive, whatever $x$ is. So after solving the quadratic in $y$, reject any root with $y \le 0$ before taking logs — that branch has no solution in $x$ at all.
Worked example 2
Had the quadratic factorised as, say, $(y-2)(y+1)=0$, the root $y=-1$ would be discarded immediately — $e^x=-1$ has no solution, since $e^x$ is never negative. You'd report only the surviving positive root. Always run this check before converting back with a log.
Type 3 — combining logs, then removing them
When several logs of the same base appear, use the laws to compress each side into a single log, then use the fact that $\log_a$ is one-to-one: if $\log_a P = \log_a Q$, then $P = Q$. This turns a log equation into an ordinary algebraic one.
Combining logs and comparing arguments can introduce extraneous roots — solutions of the simplified equation that don't satisfy the original, because they'd make some log's argument zero or negative. Every solution must be substituted back to confirm every log in the original equation stays defined. This domain check is not optional extra rigour; it's part of the method.
Worked example 3
$x=-4$ is a perfectly correct root of the quadratic — the algebra isn't wrong. It's rejected because it fails the original equation's requirements, not the simplified one. Markschemes award a method mark for finding both roots and a separate mark for correctly discarding the invalid one, so always show the domain check explicitly rather than silently dropping a root.
Type 4 — the same unknown, different bases
When $x$ appears in the exponent on both sides but the bases don't match — like $5^{x+2} = 3^{2x}$ — there's no way to compare exponents directly. Take logs of both sides anyway, apply the power law to each side separately, then collect every $x$-term on one side.
Worked example 4
Treat $\log 5$ and $\log 3$ as ordinary numbers throughout — don't try to evaluate them until the very last step. Collecting the $x$-terms works exactly like solving $5x + 10 = 6x$ for $x$; the logs are just constants riding along until you factorise $x$ out.
Don't combine logs of different bases with the product or quotient law — $\log_2 x + \log_3 y$ cannot be simplified that way. And when equating arguments after combining logs (Type 3), remember the domain check applies to the original equation's logs, not the simplified algebraic one — an extraneous root can look perfectly valid once the logs are gone.
Check yourself
Solve $\log_2 x + \log_2(x-2) = 3$, checking your answer against the domain of the original equation.
(Hint: combine into a single log, equate arguments, then check both roots against $x>2$.)
The quadratic always hands you two roots; the domain of the original logs is what decides which (if either) survive. Never skip that final check.