# How do you solve 2^(n+4)=1/32?

Apr 22, 2018

$n = \left(- \ln \frac{32}{\ln} 2\right) - 4$

#### Explanation:

Apply the natural logarithm to both sides:

$\ln \left({2}^{n + 4}\right) = \ln \left(\frac{1}{32}\right)$

Recall the logarithm quotient rule, which tells us that $\ln \left(\frac{a}{b}\right) = \ln a - \ln b .$ Then, $\ln \left(\frac{1}{32}\right) = \ln 1 - \ln 32 ,$ and as $\ln 1 = 0 ,$ we get $- \ln 32.$

Furthermore, recall the logarithm exponent rule: $\ln \left({a}^{b}\right) = b \ln a .$ So, $\ln \left({2}^{n + 4}\right) = \left(n + 4\right) \ln 2$, and we end up with

$\left(n + 4\right) \ln 2 = - \ln 32$

Solve for $n$:

$\left(n + 4\right) = - \ln \frac{32}{\ln} 2$

$n = \left(- \ln \frac{32}{\ln} 2\right) - 4$

Apr 22, 2018

$n = - 9$

#### Explanation:

$\text{note that } \frac{1}{32} = \frac{1}{2} ^ 5 = {2}^{-} 5$

$\Rightarrow {2}^{n + 4} = {2}^{-} 5 \leftarrow \textcolor{b l u e}{\text{equation to be solved}}$

$\text{Since the bases on both sides of the equation are equal}$
$\text{that is both 2, we can equate the exponents}$

$\Rightarrow n + 4 = - 5 \Rightarrow n = - 9$