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

Jun 5, 2018

$x = 0.1386 \left(4 \mathrm{dp}\right)$

#### Explanation:

${e}^{5 x} = 2$

taking natural logs of both sides

$5 x = \ln 2$

$\implies x = \frac{1}{5} \ln 2$

$x = 0.1386294361$

Jun 5, 2018

$x \approx 0.1386 \text{ to 4 dec. places}$

#### Explanation:

$\text{using the "color(blue)"laws of logarithms}$

•color(white)(x)logx^nhArrnlogx

•color(white)(x)log_b b=1

$\text{take the ln (natural log ) of both sides}$

$\ln 2 = \ln {e}^{5 x}$

$5 x {\cancel{\ln e}}^{1} = \ln 2$

$x = \ln \frac{2}{5} \approx 0.1386 \text{ to 4 dec. places}$

Jun 6, 2018

$x \approx 0.139$

#### Explanation:

$\ln$ (Natural Log) cancels with base-$e$, so we can take the natural log of both sides. We have

$\ln 2 = {\cancel{\ln e}}^{5 x}$

$\ln 2 = 5 x$

$x = \ln \frac{2}{5}$

$x \approx 0.139$

Hope this helps!