SOLUTION: Use natural logarithms to solve for x in terms of y. y = (e^x + e^−x)/(e^x − e^−x) I'm ending up with x=ln(1+y)/(y-1) but that is incorrect.

Algebra ->  Exponential-and-logarithmic-functions -> SOLUTION: Use natural logarithms to solve for x in terms of y. y = (e^x + e^−x)/(e^x − e^−x) I'm ending up with x=ln(1+y)/(y-1) but that is incorrect.       Log On


   

Question 951987: Use natural logarithms to solve for x in terms of y.
y = (e^x + e^−x)/(e^x − e^−x)
I'm ending up with x=ln(1+y)/(y-1) but that is incorrect.
Answer by josgarithmetic(39620) About Me  (Show Source):
You can put this solution on YOUR website!
y=%28e%5Ex%2Be%5E%28-x%29%29%2F%28e%5Ex-e%5E%28-x%29%29

y=%28%28e%5Ex%2Be%5E%28-x%29%29%2F%28e%5Ex-e%5E%28-x%29%29%29%28e%5Ex%2Fe%5Ex%29

y=%28e%5E%282x%29%2B1%29%2F%28e%5E%282x%29-1%29

y%28e%5E%282x%29-1%29=e%5E%282x%29%2B1

Steps up to that should make sense, and you can finish to obtain a different result than before.