Question

How do you find the inverse of `y = ln(x) + ln(x-6)`?

Answer 1

Solve for `x` using properties of logarithms and the quadratic formula and eliminate an extraneous solution to find
`f^-1(x) = 3 + sqrt(36-4e^x)/2`

Explanation:

We'll proceed under the assumption you are trying to find the inverse of the function `f(x) = ln(x) + ln(x-6)`

In general, to find the inverse of a function, a good method is to set `y = f(x)` and then solve for `x` to obtain `x = f^(-1)(y)`
(To see why this works, substitute in `f(x)` for `y` and note that the result is `f^(-1)(f(x)) = x` as desired.)

To do that here, we will need to use the following:

• `ln(a) + ln(b) = ln(ab)`
• `e^ln(a) = a`
• The quadtratic formula:
  `ax^2 + bx + c = 0 => x = (-b +-sqrt(b^2-4ac))/(2a)`

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Let `y = f(x) = ln(x) + ln(x-6)`

(note here that as we have `ln(x-6)` it must be that `x > 6`)

`=> y = ln(x(x-6)) = ln(x^2 - 6x)`

`=> e^y = e^(ln(x^2 - 6x)) = x^2 - 6x`

`=> x^2 - 6x - e^y = 0`

`=> x = (-(-6)+-sqrt((-6)^2-4(1)(-e^y)))/(2(1))`

`= (6 +-sqrt(36 + 4e^y))/2`

`= 3 +-sqrt(36 + 4e^y)/2`

But `3 - sqrt(36+4e^y)/2 < 6`, so, as noted above, we must throw it out as a possible solution for `x`

Thus

`x = 3 + sqrt(36-4e^y)/2`

Then, by our process, we have

`f^-1(x) = 3 + sqrt(36-4e^x)/2`