## Precalculus (10th Edition)

$x=0$, $x=\sqrt2$ or $x=-\sqrt2$ .
$9^x=(3^2)^x=3^{2x}$, hence the equation becomes: \begin{align*}3^{2x}=3^{x^3}.\end{align*} The base is same on the 2 sides on the equation (and it is not $1$ or $-1$), hence they will be equal if the exponents are equal. Thus \begin{align*}x^3=2x\\x^3-2x=0\\x(x^2-2)=0\end{align*} , hence $x(x+\sqrt2)(x-\sqrt2)=0.$ Hence $x=0$, $x=\sqrt2$ or $x=-\sqrt2$ by the Zero property rule.