How do you solve $4 e^{r + 5} = 29$ $4e^(r+5)=29$ ?

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Answer 1 · 2016-10-02T02:15:34

$r = ln (\frac{29}{4}) - 5 a a a$ $r=ln(29/4)-5color(white)(aaa)$ or $a a r = - 3.24214$ $color(white)(aa)r=-3.24214$

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

$\frac{4 e^{r + 5}}{4} = \frac{29}{4} a a a a$ $(4e^(r+5))/4=29/4color(white)(aaaa)$ Divide both side by 4

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

${ln e}^{r + 5} = ln (\frac{29}{4}) a a a a$ $lne^(r+5)=ln(29/4)color(white)(aaaa)$ Take the natural log of both sides

$(r + 5) ln e = ln (\frac{29}{4}) a a a$ $(r+5)lne=ln(29/4)color(white)(aaa)$ Use the log rule ${log x}^{a} = a log x$ $logx^a=alogx$

$r + 5 = ln (\frac{29}{4}) a a a a a a a$ $r+5=ln(29/4)color(white)(aaaaaaa)$ Recall that $ln e = 1$ $lne=1$

$r = ln (\frac{29}{4}) - 5 a a a a$ $r=ln(29/4)-5color(white)(aaaa)$ Subtract 5 from both sides

$r = - 3.24214 a a a a$ $r=-3.24214color(white)(aaaa)$ Use a calculator

How do you solve 4er+5=294e^(r+5)=29?