Question

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

Answer 1

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

Explanation:

`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`.

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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`.

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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]` .