How do you solve -3=log_7x?

$x = \frac{1}{343}$
${\log}_{b} p = q$ means b^q=p#
Therefore ${\log}_{7} x = - 3$ means ${7}^{- 3} = x$
$x = \frac{1}{{7}^{3}} = \frac{1}{343}$