## Calculus (3rd Edition)

$$x=±e$$
$\ln{x^4}-\ln{x^2}=2$ Since $\log{a}-\log{b}=\log{\frac{a}{b}}$, $\ln\frac{x^4}{x^2}=2$ Thus, $\ln{x^2}=2$ Raising $e$ to the power of both sides: $x^2=e^2$ Solving for $x$: $x=±e$