Given that $u = \left(\dfrac{3}{2}\right)^x$, express $9^x + 6^x = 6(4^x)$ as a quadratic equation in $u$. Hence, or otherwise, solve $9^x + 6^x = 6\bigl(4^x\bigr)$.

1. Divide the entire equation by $4^x$ to create terms with a common base.

$\frac{9^x}{4^x} + \frac{6^x}{4^x} = \frac{6(4^x)}{4^x}$

2. Simplify the fractions using laws of indices to express them in terms of $\frac{3}{2}$.

$\left(\frac{9}{4}\right)^x + \left(\frac{6}{4}\right)^x = 6$
$\left(\left(\frac{3}{2}\right)^2\right)^x + \left(\frac{3}{2}\right)^x = 6$
$\left(\left(\frac{3}{2}\right)^x\right)^2 + \left(\frac{3}{2}\right)^x = 6$

3. Substitute $u = \left(\frac{3}{2}\right)^x$ to form the quadratic equation.

$u^2 + u = 6$
$u^2 + u - 6 = 0$

1. Use the quadratic equation from part (a) to find the values of $u$.

From part (a), dividing the equation by $4^x$ gives $\left(\frac{9}{4}\right)^x + \left(\frac{6}{4}\right)^x = 6$, which simplifies to $u^2 + u - 6 = 0$.
$(u + 3)(u - 2) = 0$
$u = -3 \text{ or } u = 2$

2. Substitute $u = \left(\frac{3}{2}\right)^x$ back and reject the invalid solution.

$\left(\frac{3}{2}\right)^x = -3 \quad \text{(rejected, since } a^x > 0 \text{ for } a > 0)$
$\left(\frac{3}{2}\right)^x = 2$

3. Solve for $x$ using logarithms.

$\ln\left(\left(\frac{3}{2}\right)^x\right) = \ln 2$
$x \ln 1.5 = \ln 2$
$x = \frac{\ln 2}{\ln 1.5}$
$x \approx 1.7095 \approx 1.71 \text{ (to 3 s.f.)}$