# How do you solve 1-2e^(2x)=-19?

May 1, 2018

$x = \ln \setminus \sqrt{10}$

#### Explanation:

$1 - 2 {e}^{2 x} = - 19$

$- 2 {e}^{2 x} = - 19 - 1 = - 20$

${e}^{2 x} = - \frac{20}{- 2} = 10$

$\ln {e}^{2 x} = \ln 10$

$2 x = \ln 10$

$x = \frac{\ln 10}{2} = \ln \setminus \sqrt{10}$

Check:

$1 - 2 {e}^{2 x}$

$= 1 - 2 {e}^{2 \left(\ln \sqrt{10}\right)}$

$= 1 - 2 {e}^{\ln 10}$

$= 1 - 2 \left(10\right)$

 = -19 quad sqrt

the value is $\approx 1.151$

#### Explanation:

given $1 - 2 {e}^{2 x} = - 19 \Rightarrow - 2 {e}^{2 x} = - 20 \Rightarrow {e}^{2 x} = 10$
in general we have ${e}^{m} = k \Rightarrow {\log}_{e} k = m$
which means we have ${\log}_{e} 10 = 2 x$ and ${\log}_{e} 10 \approx 2.302$
we have $2 x = 2.302 \Rightarrow x \approx 1.151$

May 1, 2018

$x = \frac{\ln 10}{2}$
$\approx 1.1512925465$

#### Explanation:

Subtract 1 on both sides.
$- 2 {e}^{2 x} = - 20$
Divide by -2.
${e}^{2 x} = 10$
Taking the logarithm of both sides, we have:
$\ln \left({e}^{2 x}\right) = \ln 10$
Using the power rule of logarithms,
$2 x \ln \left(e\right) = \ln 10$

$\ln e = 1$ So, we have:

$2 x = \ln 10$
$x = \frac{\ln 10}{2}$