Inverse help?

Question

(1+e^x)/(1-e^x) - what is the inverse

thanks for any help

Answer 1

y=1-e^2x
x=1-e^2y
e^2y=1-x
2y=ln(1-x)
y=[ln(1-x)]/2

Answer 2

Find the inverse of y = (1 + e^x)/(1 - e^x).

First simplify the function.

y = (1 + e^x)/(1 - e^x). = 1 + 2e^x/(1 - e^x)
y = 1 + 2/{e^(-x) - 1}

To find the inverse function, switch x for y and solve for y.

x = 1 + 2/{e^(-y) - 1}
x - 1 = 2/{e^(-y) - 1}
e^(-y) - 1 = 2/(x - 1)
e^(-y) = 1 + 2/(x - 1) = {(x - 1) + 2}/(x - 1) = (x + 1)/(x - 1)
ln{e^(-y)} = ln{(x + 1)/(x - 1)}
-y = ln{(x + 1)/(x - 1)}
y = ln{(x - 1)/(x + 1)}

The inverse function is:

y = ln{(x - 1)/(x + 1)}

Looks like Helmut got it first.

Answer 3

raj is stoned.

multiplicative inverse is (1-e^x)/(1+e^x) (just switch top and bottom)

Inverse function goes as follows:
a. reduce the expression to a single occurence of x
(1+e^x)/(1-e^x)
=1+2e^x/(1-e^x)
=1+2/(e^(-x)-1)
b. do the inverse:
y = 1+2/(e^(-x)-1)
e^(-x) = 2/(y-1) +1
x= - ln( 2/(y-1) +1 )

Answer 4

let
y = (1 + e^x) / (1 - e^x)
y - ye^x = 1 + e^x
e^x + ye^x = y - 1
e^x = (y - 1)/(y + 1)
x = ln((y - 1)/(y + 1))

Exchanging y and x,
y = ln((x - 1)/(x + 1)