SOLUTION: So I have this problem y={{{(e^x)/(1-e^x)}}} for which I want to solve for x. What I tried to do was take the log of numerator and denominator separately to get x={{{log(e,y)-l

Algebra ->  Complex Numbers Imaginary Numbers Solvers and Lesson -> SOLUTION: So I have this problem y={{{(e^x)/(1-e^x)}}} for which I want to solve for x. What I tried to do was take the log of numerator and denominator separately to get x={{{log(e,y)-l      Log On


   



Question 1042521: So I have this problem y=%28e%5Ex%29%2F%281-e%5Ex%29 for which I
want to solve for x. What I tried to do was take the log
of numerator and denominator separately to get x=log%28e%2Cy%29-log%28e%2C%281-y%29%29
Combining that I get log%28e%2C%28y%2F1-y%29%29 but apparently the answer was log%28e%2C%28y%2F1%2By%29%29.
So my question is why did the sign switch from a + to a -.
Thanks

Answer by Edwin McCravy(20060) About Me  (Show Source):
You can put this solution on YOUR website!


I use "ln" instead of "loge.

You took natural logs of both sides too soon.
Wait until you've solved for ex.

y%22%22=%22%22%28e%5Ex%29%2F%281-e%5Ex%29

 First multiply both sides by 1-ex

y%281-e%5Ex%29%22%22=%22%22e%5Ex

Distribute

y-ye%5Ex%22%22=%22%22e%5Ex

Isolate the terms in ex:

y%22%22=%22%22ye%5Ex%2Be%5Ex

Factor out ex on the right:

y%22%22=%22%22e%5Ex%2A%28y%2B1%29

Divide both sides by y+1

y%2F%28y%2B1%29%22%22=%22%22e%5Ex

Take natural logs of both sides:

ln%28y%2F%28y%2B1%29%29%22%22=%22%22x

x%22%22=%22%22ln%28y%2F%28y%2B1%29%29

Edwin