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

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Answer 1 · 2015-12-13T00:23:56

$x= ln(1498)/2$

We will use the following properties of logarithms:

$3000/(2+e^(2x)) = 2$

$=> 3000 = 2(2+e^(2x))=4 + 2e^(2x)$

$=> 2996 = 2e^(2x)$

$=> 1498 = e^(2x)$

$=> ln(1498) = ln(e^(2x)) = 2xln(e) = 2x(1) = 2x$

$=> x = ln(1498)/2$

How do you solve 3000/(2+e^(2x))=23000/(2+e^(2x))=2?