Question

h(x) = log (6x+5/7x-4)

Find a formula for the inverse function h^-1(x)

h^-1(x) =

My work so far h(x) = log(6x+5/7x-4)
y=log(6x+5/7x-4)

x=log(6y+5/7y-4)

7xy-4x= log(6y+5)

how would I solve from here. Would I solve for y then do the natural log of both sides or do the natural log first then solve for y?

Answer 1

First, you have to note that since log(z) is defined only for z>0, we must have

x < -4/5 or x > 4/7

In that domain, there is an inverse.

You made an error in your last line, which is where the trouble arises. It should be

x = log (6y+5)/(7y-4)
e^x = (6y+5)/(7y-4)
e^x(7y-4) = 6y+5
7e^x y - 4e^x = 6y+5
(7e^x-6)y = 4e^x+5
y = (4e^x+5)/(7e^x-6)