# How do you solve 2^(x+8)=5?

${2}^{x + 8} = 5$
$= {.2}^{x} \cdot {2}^{8} = 5$
$\implies {2}^{x} = \frac{5}{2} ^ 8$
$\implies x \log 2 = \left(\log 5 - 8 \log 2\right)$
$x = \log \frac{5}{\log} 2 - 8$