# How do you solve 4^(x+5)=0.5?

Apr 5, 2016

$x = - \frac{11}{2}$

#### Explanation:

${4}^{x + 5} = 0.5$

First apply logarithms because $\textcolor{b l u e}{a = b \implies \ln a = \ln b , \mathmr{if} a , b > 0}$

$\left(x + 5\right) \ln 4 = \ln \left(0.5\right)$

$\left(x + 5\right) \ln \left({2}^{2}\right) = \ln \left({2}^{-} 1\right)$

$\left(x + 5\right) \cdot 2 \cdot \ln \left(2\right) = - \ln \left(2\right)$

ln(2) is a constant, so you can divide the expression by it

$\left(x + 5\right) \cdot 2 = - 1$

$2 x + 10 = - 1$

$2 x = - 11$

$x = - \frac{11}{2}$