Solution — 2016 Question 2 Exponential & Logarithmic Functions| Quadratic Equations & Inequalities

\begin{matrix} 2 e^{2 x} & \geq 9 - 3 e^{x} \end{matrix}

Let $u = e^{x} .$ Since $e^{2 x} = u^{2},$ substituting into the inequality,

\begin{matrix} 2 u^{2} & \geq 9 - 3 u \\ 2 u^{2} + 3 u - 9 & \geq 0 \\ (u + 3) (2 u - 3) & \geq 0 \end{matrix}

$u \leq - 3 or u \geq \frac{3}{2}$

Since $u = e^{x} > 0,$ we reject $u \leq - 3$

\begin{matrix} e^{x} & \geq \frac{3}{2} \\ \ln e^{x} & \geq \ln \frac{3}{2} \\ x & \geq \ln \frac{3}{2} ∎ \end{matrix}