Inequalities: CJC 2024 Promo Q2

$$\begin{aligned} \frac{3x - 7}{(x+3)(x-2)} - 1 &= \frac{3x - 7 - (x+3)(x-2)}{(x+3)(x-2)}\\ &= \frac{-x^2-x+6+3x-7}{(x+3)(x-2)}\\ &= \frac{-x^2 + 2x - 1}{(x+3)(x-2)}. \end{aligned}$$ $$\begin{aligned} \frac{3x - 7}{(x+3)(x-2)} &< 1\\ \frac{3x - 7}{(x+3)(x-2)} - 1 &< 0\\ \frac{-x^2 + 2x - 1}{(x+3)(x-2)} &< 0\\ \frac{x^2 - 2x + 1}{(x+3)(x-2)} &> 0\\ \frac{(x - 1)^2}{(x+3)(x-2)} &> 0. \end{aligned}$$

$$\begin{aligned} \therefore\ x&<-3\text{ or }x>2. \end{aligned}$$ Replace $x$ by $\mathrm{e}^x$, $$\begin{aligned} \mathrm{e}^x&<-3\ \text{(reject $\because\mathrm{e}^x>0$) or }\mathrm{e}^x>2\\ x&>\ln2 \end{aligned}$$