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

1 Answer
Dec 4, 2015

Answer:

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)#

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#