Answer
$x=-10$
Work Step by Step
$\left(\frac{1}{27}\right)^{4-x}=9^{2x-1}$
Rewriting $27$ as $3^{3}$ and $9$ as $3^{2}$, we have
$\left(\frac{1}{3^{3}}\right)^{4-x}=(3^{2})^{2x-1}$
$\implies (3^{-3})^{4-x}=3^{4x-2}$
$\implies 3^{-12+3x}=3^{4x-2}$
Equating the exponents, we get
$-12+3x=4x-2$
$-12+2=4x-3x$
$x=-10$
Check:
$\left(\frac{1}{27}\right)^{4-(-10)}=(3^{-3})^{14}=3^{-42}$
$9^{2(-10)-1}=(3^2)^{-21}=3^{-42}$, which is true.
