A Hidden Quadratic in Logarithms

An equation with $(\log x)^2$ in it is a quadratic in disguise. Substitute $u = \log_3 x$, solve the quadratic, then convert back. Watch the power law: $\log_3 x^2 = 2\log_3 x$, not $(\log_3 x)^2$.

Solve $(\log_3 x)^2 - \log_3 x^2 = 3$. [5]

The working

Step 1, apply the power law to the second term: $\log_3 x^2 = 2\log_3 x, \quad \text{so the equation is } (\log_3 x)^2 - 2\log_3 x = 3. \quad \text{(M1)}$

Step 2, substitute $u = \log_3 x$: $u^2 - 2u = 3 \implies u^2 - 2u - 3 = 0 \quad \text{(M1)}$

Step 3, factorise and solve for $u$: $(u - 3)(u + 1) = 0 \implies u = 3 \ \text{or} \ u = -1 \quad \text{(A1)}$

Step 4, convert back using $u = \log_3 x \Leftrightarrow x = 3^u$: $x = 3^3 = 27 \qquad \text{or} \qquad x = 3^{-1} = \tfrac{1}{3} \quad \text{(A1, A1)}$

Where the marks are won and lost

• Distinguish $\log_3 x^2 = 2\log_3 x$ from $(\log_3 x)^2 = u^2$. Confusing them wrecks the quadratic.
• After solving for $u$, convert both values back to $x$. A negative $u = -1$ is fine and gives $x = \frac13$ (still positive, so valid).
• Bring the equation to $u^2 - 2u - 3 = 0$ before factorising.

Common mistakes

• Treating $\log_3 x^2$ as $(\log_3 x)^2$ and getting the wrong equation.
• Rejecting $u = -1$ as “impossible” (it gives a valid $x = \frac13$; it’s $x$ that must be positive, not $u$).
• Forgetting to convert $u$ back to $x$.

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