Solve log₃(x) + log₃(x-2) = 1

Question

Solve for $x$ : $\log_3(x) + \log_3(x-2) = 1$

Solution — Step by Step

Before solving, we must note that logarithms are only defined for positive arguments.

So we need:

$x > 0$ (for $\log_3 x$ to be defined)
$x - 2 > 0 \Rightarrow x > 2$ (for $\log_3(x-2)$ to be defined)

The combined restriction is $x > 2$ . Any solution must satisfy this.

Using the logarithm product rule: $\log_b M + \log_b N = \log_b(MN)$

\log_3(x) + \log_3(x-2) = \log_3[x(x-2)]

So the equation becomes:

\log_3[x(x-2)] = 1

If $\log_b A = c$ , then $A = b^c$ .

x(x-2) = 3^1 = 3

x^2 - 2x = 3

x^2 - 2x - 3 = 0

Factor: $x^2 - 2x - 3 = (x-3)(x+1) = 0$

So $x = 3$ or $x = -1$ .

We established that $x > 2$ .

$x = 3$ : $3 > 2$ ✓ — valid solution
$x = -1$ : $-1 > 2$ ✗ — rejected (also $\log_3(-1)$ is undefined)

The solution is $x = 3$ .

Verification: $\log_3(3) + \log_3(3-2) = 1 + \log_3(1) = 1 + 0 = 1$ ✓

Why This Works

The logarithm product rule ( $\log A + \log B = \log AB$ ) is the key tool here — it converts a sum of logs into one log, which we can then write as an exponential equation. This converts the transcendental equation into a manageable polynomial equation.

The domain check step is non-negotiable. Quadratic equations can produce two mathematical solutions, but logarithms require positive arguments. Without checking, we would incorrectly include $x = -1$ .

Notice that $x = -1$ doesn’t just give a negative argument — it gives $\log_3(-1)$ which is undefined in real numbers. The extraneous root appears because squaring (or multiplying) operations in equation-solving don’t preserve domain restrictions — we must always re-check.

Alternative Method

We can also solve by noticing $3^1 = 3$ . Write 3 as $3 \times 1$ : we need two numbers $x$ and $x-2$ such that their product is 3. We’re looking for $x$ and $x-2$ that multiply to 3. Try $x = 3$ : $3 \times 1 = 3$ ✓. This doesn’t work as a general method but gives a quick mental check for “nice” problems.

A more systematic alternative: $\log_3(x-2) = 1 - \log_3 x = \log_3 3 - \log_3 x = \log_3(3/x)$

Then $x-2 = 3/x \Rightarrow x(x-2) = 3$ — same quadratic as before.

For JEE and CBSE Class 11 logarithm problems, the sequence is always: (1) state domain restrictions, (2) use log rules to combine into single log, (3) convert to exponential form, (4) solve the resulting polynomial, (5) check each solution against domain. Skipping step 1 and step 5 is the most common reason for losing full marks. Examiners specifically check for the domain condition in marking schemes.

Common Mistake

The most frequent error is accepting $x = -1$ as a valid solution without domain checking. After solving the quadratic $(x-3)(x+1) = 0$ , many students write “x = 3 or x = -1” as the final answer. This is wrong — $x = -1$ is an extraneous root introduced when we applied the product rule and formed the equation $x(x-2) = 3$ (we implicitly allowed any $x$ , even negative ones). Always check both solutions in the original logarithmic equation, not just the quadratic.

Question

Solution — Step by Step

Why This Works

Alternative Method

Common Mistake

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