How do you find the inverse of # f(x) = ln(4 - 7x) + ln(-7 - 5x)#?

Let #x = f^-1(x)#

Substitute #f^-1(x)# everywhere there is an x:

#f(f^-1(x)) = ln(4 - 7f^-1(x)) + ln(-7 - 5f^-1(x))#

Use the property of inverses #f(f^-1(x)) = x#:

#x = ln(4 - 7f^-1(x)) + ln(-7 - 5f^-1(x))#

Use the identity #ln(a) + ln(b) = ln(ab)#

#x = ln((4 - 7f^-1(x)(-7 - 5f^-1(x)))#

Use the inverse of the natural logarithm on both sides:

#e^x = (4 - 7f^-1(x)(-7 - 5f^-1(x))#

Use the F.O.I.L. method to perform the multiplication:

#e^x = -28 - 20f^-1(x) + 49f^-1(x) + 35(f^-1(x))^2#

This is a quadratic.

#35(f^-1(x))^2 + 29f^-1(x) - 28 - e^x = 0#

At this point, we must declare that there is no inverse, because the quadratic formula yields two equations that returns two values for any given x and an inverse must not do this.

Let's proceed so that you can see the issue

Use the quadratic formula, #x = {-b +-sqrt(b^2 - 4(a)(c))}/{2a}#

#f^-1(x) = {-29 + sqrt(29^2 - 4(35)(-28-e^x))}/{2(35)}#

AND

#f^-1(x) = {-29 - sqrt(29^2 - 4(35)(-28-e^x))}/{2(35)}#

This is contrary to the purpose of an inverse function.

The purpose of an inverse function is, for any #y# returned from the # f(x)#, give you back a single value of #x# that created the #y#; this returns two values, therefore, no inverse.