A cone has a volume of $(24 + 5\sqrt{3})\pi\text{ cm}^3$ and a base radius of $(2 + \sqrt{3})\text{ cm}$. Find its height in the form of $(p + q\sqrt{3})$, where $p$ and $q$ are integers. Solve the equation $e^x(3e^x - 2) = 5$.

1. Expand the equation to form a quadratic equation in $e^x$

$3(e^x)^2 - 2e^x - 5 = 0$

2. Factorise the quadratic equation and solve for $e^x$

Let $y = e^x$.
$3y^2 - 2y - 5 = 0$
$(3y - 5)(y + 1) = 0$
$y = \frac{5}{3} \text{ or } y = -1$
$e^x = \frac{5}{3} \text{ or } e^x = -1$

3. Solve for $x$ and reject the invalid solution

Since $e^x > 0$ for all real $x$, $e^x = -1$ has no real solution (reject).
$e^x = \frac{5}{3}$
$x = \ln\left(\frac{5}{3}\right) \approx 0.511 \text{ (to 3 s.f.)}$