How do you solve log_5 x=-3?

$x = \frac{1}{125} = 0.008$.
Inversely, $x = {5}^{- 3} = \frac{1}{5} ^ 3 = \frac{1}{125} = 0.008$.
Note that if ${\log}_{a} x = b , x = {a}^{b}$ and, inversely, if $x = {a}^{b} , {\log}_{a} x = b$..