# How do you solve e^(2x) = ln(4 + e)?

May 27, 2016

I found: $x = \frac{\ln \left(\ln \left(4 + e\right)\right)}{2} = 0.322197$

#### Explanation:

We could call the constant $\ln \left(4 + e\right)$, say, $k$ and write:
$e \left(^ 2 x\right) = k$
take the natural log of both sides:
$\ln \left({e}^{2 x}\right) = \ln \left(k\right)$
and:
$2 x = \ln \left(k\right)$ where $\ln$ and $e$ eleiminated one another;
finally:
$x = \ln \frac{k}{2}$
Now:
if we can use a calculator we have that:
$k = \ln \left(4 + e\right) = 1.90483$
so that $x$ becomes:
$x = \frac{\ln \left(\ln \left(4 + e\right)\right)}{2} = 0.322197$