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

May 5, 2015

You can try by taking the logarithm (of base $4$) of both sides:
${\log}_{4} \left({4}^{x - 1}\right) = {\log}_{4} \left(3\right)$
${\log}_{4}$ and ${4}^{}$ cancel out, so you get:
$x - 1 = {\log}_{4} \left(3\right)$ and
$x = 1 + {\log}_{4} \left(3\right)$

This can be a little tricky to solve unless you decide to change base and can use a calculator (or tables):
${\log}_{4} \left(3\right) = \frac{\ln 3}{\ln 4} \approx 0.8$ (using natural logarithms)
So that:
$x = 1.8$