Answer
$x=\pm1$
Work Step by Step
$x^{2}2^{x}-2^{x}=0$
Take out common factor $2^{x}$:
$2^{x}(x^{2}-1)=0$
Set both factor equal to $0$ and solve each individual equation:
$2^{x}=0$
Since no value of $x$ makes this first equation true, it has no solution. Let's move on to the next one:
$x^{2}-1=0$
Take the $-1$ to the right side of the equation:
$x^{2}=1$
Take the square root of both sides:
$\sqrt{x^{2}}=\sqrt{1}$
$x=\pm1$
