## Precalculus (10th Edition)

$x=-\frac{3}{2}$
$4^{x-3}=8^{2x}\\(2^2)^{x-3}=(2^3)^{2x}\\2^{2(x-3)}=2^{3(2x)}\\2^{2x-6}=2^{6x}$. We know that if $x\ne1,\ne-1$ then $x^a=x^b\longrightarrow a=b$. Hence here: $2x-6=6x\\2x=6x+6\\-4x=6\\x=-\frac{3}{2}$