Solution — 2023 Question 1 Exponential & Logarithmic Functions| Quadratic Equations & Inequalities

\begin{matrix} 3 e^{x} & \geq 4 e^{- x} - 4 \\ 3 e^{x} & \geq \frac{4}{e^{x}} - 4 \end{matrix}

Let $u = e^{x} .$ Substituting into the inequality,

$3 u \geq \frac{4}{u} - 4$

Multiplying both sides by $u$ (since $u = e^{x} > 0$ ),

\begin{matrix} 3 u^{2} & \geq 4 - 4 u \\ 3 u^{2} + 4 u - 4 & \geq 0 \\ (u + 2) (3 u - 2) & \geq 0 \end{matrix}

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

Since $u = e^{x} > 0,$ we reject $u \leq - 2$

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