# How do you solve log_6(4x+4)=log_6 64?

Nov 22, 2016

$x = 15$

#### Explanation:

${\log}_{6} \left(4 x + 4\right) = {\log}_{6} 64$

Hence ${\log}_{6} \left(4 x + 4\right) - {\log}_{6} 64 = 0$

or ${\log}_{6} \left(\frac{4 x + 4}{64}\right) = 0 = {\log}_{6} 1$

or $\frac{4 x + 4}{64} = 1$ i.e. $4 x + 4 = 64$

and $x = \frac{64 - 4}{4} = \frac{60}{4} = 15$