what is the inverse function of y=(1+e^x)/(1-e^x)?

Question

Could I get a good explanation for that? Thanx

Answer 1

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

Answer 2

It f(x) = (1 + e^x)/(1 - e^x), we are looking for a function g(x) such that f(g(x)) = x. But
f(g(x)) = [1 + e^(g(x))]/[1 - e^(g(x)] = x leads to 1 + e^(g(x)) = x - x*e^(g(x)) so
[e^(g(x))]*(1 + x) = x - 1, and e^(g(x)) = (x - 1)/(x + 1). Therefore, g(x) = ln((x - 1)/(x + 1)).

As a check, you can verify that g(f(x)) = x, as required.

Answer 3

If y=(1 + e^x)/(1 - e^x)

Multiply both sides by (1 - e^x)

(1 - e^x)y = (1 + e^x)

y - (e^x)y = 1 + e^x

y = 1 + e^x + (e^x)y

y = 1 +(1 + y)e^x

y - 1 = (1 + y)e^x

(y - 1)/(y + 1) = e^x

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

x = ln(y - 1) - ln(y + 1)

Answer 4

first, change y and x, around, so the expression is in terms of x

x = (1+e^y)/(1-e^y)

then express in terms of y

x(1-e^y) =(1+e^y)

x - ex^y = 1+ e^Y

x - 1 = xe^y +e^Y

x - 1 = e^y(x + 1)

e^y = (x - 1)/(x+1)

then take logs of both sides..if you want...

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