## Precalculus (10th Edition)

$x=6.$
The base is same on the 2 sides on the equation (and it is not $1$), hence they will be equal if the exponents are equal. Hence $\log_4{(x+2)}=\log_4{8}$ if and only if $8=x+2$, hence after subtracting 2 from both sides: $x+2-2=8-2\\x=6.$