# How do you solve 4e^(r+5)=29?

Oct 2, 2016

$r = \ln \left(\frac{29}{4}\right) - 5 \textcolor{w h i t e}{a a a}$ or $\textcolor{w h i t e}{a a} r = - 3.24214$

#### Explanation:

$4 {e}^{r + 5} = 29$

$\frac{4 {e}^{r + 5}}{4} = \frac{29}{4} \textcolor{w h i t e}{a a a a}$ Divide both side by 4

${e}^{r + 5} = \frac{29}{4}$

$\ln {e}^{r + 5} = \ln \left(\frac{29}{4}\right) \textcolor{w h i t e}{a a a a}$ Take the natural log of both sides

$\left(r + 5\right) \ln e = \ln \left(\frac{29}{4}\right) \textcolor{w h i t e}{a a a}$ Use the log rule $\log {x}^{a} = a \log x$

$r + 5 = \ln \left(\frac{29}{4}\right) \textcolor{w h i t e}{a a a a a a a}$ Recall that $\ln e = 1$

$r = \ln \left(\frac{29}{4}\right) - 5 \textcolor{w h i t e}{a a a a}$ Subtract 5 from both sides

$r = - 3.24214 \textcolor{w h i t e}{a a a a}$ Use a calculator