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

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

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Answer 1 · 2017-01-14T17:05:49

The answer is $x = \frac{1}{2} ln (1498) = ln (\sqrt{1498}) \approx 3.65594$ $x=1/2 ln(1498)=ln(sqrt(1498))approx 3.65594$

Explanation:

First, multiply both sides by $2 + e^{2 x}$ $2+e^(2x)$ and divide both sides by 2 to get $2 + e^{2 x} = 1500 .$ $2+e^(2x)=1500.$ Subtracting 2 from both sides leads to $e^{2 x} = 1498$ $e^(2x)=1498$ . Now take the natural logarithm of both sides and then divide both sides by 2 to get the final answer:

$x = \frac{1}{2} ln (1498) = ln (\sqrt{1498}) \approx 3.65594$ $x=1/2 ln(1498)=ln(sqrt(1498))approx 3.65594$ .

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