# How do you solve  log_12 (x + 1) – log_12 4 = log_12 16?

May 10, 2015

Here you need to remember some rules of logarithms!
Consider that:
$\log a - \log b = \log \left(\frac{a}{b}\right)$
So you can write:
${\log}_{12} \left(\frac{x + 1}{4}\right) = {\log}_{12} \left(16\right)$
Also remember that ${a}^{{\log}_{a} \left(x\right)} = x$
So use $12$ to eliminat the logs as:
${12}^{{\log}_{12} \left(\frac{x + 1}{4}\right)} = {12}^{{\log}_{12} \left(16\right)}$
which gives you:
$\frac{x + 1}{4} = 16$
And:
$x = 63$