Without using a calculator, solve
$x&lt;\frac{3}{x-2}$.[3]
Hence solve
${{\mathrm{e}}^{-x}}&lt;\frac{3}{{{\mathrm{e}}^{-x}}-2}$.[2]

${{\mathrm{e}}^{-x}}<\frac{3}{{{\mathrm{e}}^{-x}}-2}$.

Inequalities - General

Inequalities

Multiply both sides by \$-1\$ (reverse inequality):

\$0 > \\dfrac{(x+1)(x-3)}{x-2}\$, where \$x \\neq 2\$

Using number line with critical values \$x = -1, 2, 3\$:

\$\\therefore \\{x \\in \\mathbb{R}: x < -1 \\text{ or } 2 < x < 3\\}\$

Part (b) - Hence solve e⁻ˣ < 3/(e⁻ˣ - 2)

\$e^{-x} < -1\$ or \$2 < e^{-x} < 3\$

Since \$e^{-x} > 0\$ always, \$e^{-x} < -1\$ is rejected.