How do you solve $\frac{3000}{2 + e^{2 x}} = 2$ $3000/(2+e^(2x))=2$ ?

Question

2128 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-12-13T00:23:56

$x = \frac{ln (1498)}{2}$ $x= ln(1498)/2$

We will use the following properties of logarithms:

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

$\Rightarrow 3000 = 2 (2 + e^{2 x}) = 4 + 2 e^{2 x}$ $=> 3000 = 2(2+e^(2x))=4 + 2e^(2x)$

$\Rightarrow 2996 = 2 e^{2 x}$ $=> 2996 = 2e^(2x)$

$\Rightarrow 1498 = e^{2 x}$ $=> 1498 = e^(2x)$

$\Rightarrow ln (1498) = ln (e^{2 x}) = 2 x ln (e) = 2 x (1) = 2 x$ $=> ln(1498) = ln(e^(2x)) = 2xln(e) = 2x(1) = 2x$

$\Rightarrow x = \frac{ln (1498)}{2}$ $=> x = ln(1498)/2$

How do you solve 30002+e2x=23000/(2+e^(2x))=2?