# How do you solve 4^(x-1)=3?

The value of $x = 1 + \log \frac{3}{\log} 4$
$\log {4}^{x - 1} = \log 3 \implies \left(x - 1\right) \log 4 = \log 3 \implies x - 1 = \log \frac{3}{\log} 4 \implies x = 1 + \log \frac{3}{\log} 4$