How do you solve $40 / (1+e^(-2x)) = 8$ ?

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Answer 1 · 2015-07-07T16:45:19

When solving a complicated equation, try and undo all the things being done to the variable, and in the reverse order!

In this case, multiply through by the denominator and get

$40 = 8 * (1 + e^(-2x))$

Now isolate the $x,$ step-by-step:

$5 = 1 + e^(-2x)$

$4 = e^(-2x)$

$4 = 1/(e^(2x)) <=> e^(2x) = 1/4$

This means that you get

$2x = ln(1/4) = -ln 4$ ;

$x = -ln 2$ , or $x ~= -0.693$ .

I left out a couple of steps near the end to make you think!

How do you solve 40 / (1+e^(-2x)) = 840 / (1+e^(-2x)) = 8?