Question

Solve the following equations.
$\frac{3^x+3-x}{3^x-3-x}=2.$

Answer 1

$\frac{3^x+3^{-x}}{3^x-3^{-x}}=2$

Let, $b=3^x⟹3^{-x}=\frac{1}{b}$

We get-

$\frac{b+\frac{1}{b}}{b-\frac{1}{b}}=2$

$⟹\frac{b^2+1}{b^2-1}=2$

$⟹b^2+1=2b^2-2$

$⟹b^2=3⟹b=\sqrt{3}$

$⟹3^x=\sqrt{3}$ since this can't be negative hence $b\ne -\sqrt{3}$

$⟹x=\frac{1}{2}$