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!