Solving Logarithmic Equations

What this method is for

A logarithmic equation is any equation where the unknown sits inside a logarithm. This method turns such an equation into an ordinary one you already know how to solve, usually linear or quadratic, by peeling the logarithms away safely.

Two moves do all the work. If you can reduce the equation to one logarithm equalling another of the same base, the arguments must be equal.

If you can reduce it to one logarithm equalling a plain number, you rewrite it in index form.

equate arguments

\log_a M = \log_a N \;\Rightarrow\; M = N

index form

\log_a x = b \;\Rightarrow\; x = a^{b}

Because taking a logarithm requires a positive argument, some algebraic solutions must be rejected, checking is part of the method, not an optional extra.

When to reach for it

Use this method whenever the variable appears inside a $\log$ , such as $\log_2 x$ , $\log_3(x-1)$ , or a sum of several logarithms set equal to a number. The tell-tale signs are two or more logarithm terms that can be merged, or a single logarithm sitting alone on one side of an equals sign.

If the logarithms in the equation carry different bases, convert them to a common base first, the laws that merge logs only work within one base. If instead the unknown is in an exponent rather than inside a log, that is an index equation; you take logarithms of both sides to bring the power down, which is a related but different technique.

The steps

1
Same base
Make sure every logarithm uses the same base; convert with the change-of-base rule if they do not.
2
Combine the logs
Use the product and quotient laws to collapse several logarithms into a single logarithm on each side.
3
Remove the logarithm
If it is log = log, equate the arguments. If it is log = number, rewrite in index form using $\log_a x = b \Rightarrow x = a^{b}$ .
4
Solve the equation
Solve the resulting linear or quadratic equation for the unknown.
5
Check every root
Substitute each answer back: the argument of every original logarithm must be positive. Reject any value that makes an argument zero or negative.

Worked example

Q1[4 marks]

Solve $\log_2 x + \log_2(x-2) = 3$ .

Show worked solution

Both terms share base $2$ , so combine them with the product law.

\log_2\left[x(x-2)\right] = 3

The right-hand side is a plain number, so rewrite in index form using $\log_a x = b \Rightarrow x = a^{b}$ :

x(x-2) = 2^{3} = 8

Expand and bring everything to one side to form a quadratic equation.

x^{2} - 2x - 8 = 0

Factorise the quadratic:

(x-4)(x+2) = 0 \;\Rightarrow\; x = 4 \;\text{or}\; x = -2

Check each root. The original equation contains $\log_2 x$ and $\log_2(x-2)$ , so both $x$ and $x-2$ must be positive.

For $x = -2$ , $\log_2(-2)$ is undefined, so reject $x=-2$ . For $x = 4$ : $\log_2 4 + \log_2 2 = 2 + 1 = 3$ , which matches.

x = 4

So the only solution is $x = 4$ . The rejected root is not a mistake, it is exactly why the checking step matters.

Common pitfalls

Skipping the check, so a root that makes a logarithm's argument negative is wrongly kept.
Adding logarithms as if $\log_a x + \log_a(x-2) = \log_a(2x-2)$ ; the product law gives $\log_a[x(x-2)]$ , not a sum inside.
Converting $\log_a x = b$ the wrong way round, writing $x = b^{a}$ instead of $x = a^{b}$ .
Trying to equate arguments while a stray number still sits on one side, reduce to log = log or log = number first.
Leaving different bases in the same equation instead of converting to a common base.

How a teacher helps

The step students most often drop is the final check, and it is the step that decides whether a perfectly good quadratic earns full marks or loses one to an invalid root. In one-to-one lessons our teachers make the check a reflex: after every logarithmic equation you learn to glance back at each argument and ask 'is this positive?'

We also watch the index-form conversion closely, because writing $x = b^{a}$ instead of $x = a^{b}$ is a quiet, costly slip. Every teacher on spmaddmath.com.my is experienced.

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Frequently asked questions

Why do some solutions to a logarithmic equation get rejected?

A logarithm is only defined for a positive argument. When you solve the underlying quadratic you may get a value that makes one of the original logarithms take a zero or negative argument.

That value is not a genuine solution, so you reject it, which is why substituting back is essential.

When do I equate arguments and when do I use index form?

If the equation reduces to $\log_a M = \log_a N$ (a logarithm on each side, same base), equate the arguments: $M = N$ . If it reduces to $\log_a x = b$ (a logarithm equal to a plain number), rewrite in index form: $x = a^{b}$ .

What if the equation has logarithms with different bases?

Convert them all to one base first using $\log_a b = \frac{\log_c b}{\log_c a}$ . The product and quotient laws that let you merge logarithms only apply when every term shares the same base.

Do I get marks for the working if my final answer is wrong?

Yes. SPM Add Math uses analytic marking, so method marks are awarded for correctly combining the logarithms, converting to index form, and solving the quadratic.

Showing each step clearly protects those marks even if an arithmetic slip changes the final value.

Source:SRC-DSKP-EN

Written by the spmaddmath.com.my editorial team.· Last updated 5 September 2026

How to solve logarithmic equations