# How do you solve 3000/(2+e^(2x))=2?

Dec 13, 2015

$x = \ln \frac{1498}{2}$

#### Explanation:

We will use the following properties of logarithms:

• $\ln \left({a}^{x}\right) = x \ln \left(a\right)$

• ${\log}_{a} \left(a\right) = 1$

$\frac{3000}{2 + {e}^{2 x}} = 2$

$\implies 3000 = 2 \left(2 + {e}^{2 x}\right) = 4 + 2 {e}^{2 x}$

$\implies 2996 = 2 {e}^{2 x}$

$\implies 1498 = {e}^{2 x}$

$\implies \ln \left(1498\right) = \ln \left({e}^{2 x}\right) = 2 x \ln \left(e\right) = 2 x \left(1\right) = 2 x$

$\implies x = \ln \frac{1498}{2}$