How do you find the inverse of #y=ln(x/(x-1))#?

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

First of all, let's establish your domain:

• the denominator of the fraction isn't allowed to be equal to #0#: # x - 1 != 0 => x != 1#
• the logarithmic term needs to be greater than #0#. This is the case if both numerator and denominator are positive or if they are both negative.

As the numerator and the denominator have a difference of #1#, you can see that they can only have different signs for #0 <= x <= 1#.

Thus, your domain is #x !in [0,1] # or, if you prefer a different notation,

#x < 0# or #x > 1#.

=============================

Now let's start with finding the inverse.

At first, swap #y# and #x#:

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

Now, your goal is to solve this for #y#.
To do so, the first step would be to "get rid" of the #ln# term.

The inverse function for #ln x # is #e^x#, and both #ln (e^x) = x# and #e^(ln x) = x# hold. Thus, you can apply the #e^x# function on both sides of the equation:

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

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

Now, you would prefer to have just one #y# and not two of them. There is a way to achieve this:

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

To procede, take the reciprocal on both sides:

#1 / e^x = 1 - 1 / y#

Compute #-1# on both sides:

#-1 + 1 / e^x = - 1 /y#

Multiply with #-1# on both sides:

# 1 - 1 / e^x = 1 / y#

Finally, take the reciprocal on both sides again:

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

To avoid double fractions, this can be rephrased in:

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

This function is defined for all #x# except #x = 0# since #e^0 - 1 = 1 - 1 = 0#.

=============================

Thus, your inverse function is

#f^(-1)(x) = e^x/ (e^x - 1)#

It has the domain #x != 0# and its range is the domain of the original function: #x !in [0, 1]# .