A Hidden Quadratic in an Exponential Equation

When an exponential equation has three terms and one exponent is double another, it is a quadratic in disguise. The substitution $y = 3^x$ makes it obvious. Solve the quadratic, then convert each value of $y$ back to $x$, and reject any impossible $y$.

Solve $3^{2x} - 10\left(3^x\right) + 9 = 0$. [5]

The working

Step 1, spot the structure. Note $3^{2x} = \left(3^x\right)^2$. Let $y = 3^x$: $y^2 - 10y + 9 = 0 \quad \text{(M1)}$

Step 2, factorise the quadratic in $y$: $(y - 1)(y - 9) = 0 \;\Rightarrow\; y = 1 \quad \text{or} \quad y = 9 \quad \text{(A1)}$

Both are positive, so both are valid values of $3^x$.

Step 3, convert back with $3^x = y$: $3^x = 1 \;\Rightarrow\; x = 0 \quad (\text{since } 3^0 = 1) \quad \text{(M1, A1)}$ $3^x = 9 = 3^2 \;\Rightarrow\; x = 2 \quad \text{(A1)}$

So $x = 0$ or $x = 2$.

Where the marks are won and lost

• The substitution is the method mark. Recognising $3^{2x} = (3^x)^2$ (not $3^x \times 2$ or $9^x$ mishandled) is what unlocks the quadratic.
• Convert every root back. Stopping at $y = 1, 9$ leaves the question unanswered; the variable was $x$, not $y$.
• $3^x = 1$ gives $x = 0$, a value students often miss because it “looks like nothing”. Any base to the power $0$ is $1$.

Common mistakes

• Writing $3^{2x} = 2 \cdot 3^x$ instead of $(3^x)^2$.
• Solving for $y$ and forgetting to find $x$.
• Discarding $3^x = 1$ or misreading it as no solution.

Full method: Solving Log & Exponential Equations notes. Topic home: Logs & Exponentials pillar.