# How do you solve  log x = log 3 + 2 log x?

$x = {3}^{-} 1 = \frac{1}{3}$
$\implies 2 \log x - \log x = - \log 3$
$\implies \log x = \log {3}^{-} 1$
$x = {3}^{-} 1 = \frac{1}{3}$