## Precalculus: Mathematics for Calculus, 7th Edition

$x=7$
$5^{4-x}=\dfrac{1}{125}$ Rewrite $125$ as $5^{3}$: $5^{4-x}=\dfrac{1}{5^{3}}$ Take $5^{3}$ to the numerator by changing the sign of its exponent: $5^{4-x}=5^{-3}$ If $5^{4-x}=5^{-3}$, then $4-x=-3$ $4-x=-3$ Solve for $x$: $-x=-3-4$ $-x=-7$ $x=7$