Find the inverse of the function? : h(x) = log((x+9)/(x−6))

2 Answers
Sep 18, 2017

h^(-1)(x) = 3 ( (2e^x + 3) / (e^x - 1) )

Explanation:

We have:

h(x) = log((x+9)/(x−6))

To find h^(-1)(x) we need to rearrange to the equation to the for x=f(h).

Writing as:

h = log((x+9)/(x−6))

:. (x+9)/(x−6) = e^h

:. x+9 = (x−6)e^h

:. x+9 = xe^h−6e^h

:. xe^h - x = 6e^h + 9

:. x(e^h - 1) = 3(2e^h + 3)

:. x = 3 ( (2e^h + 3) / (e^h - 1) )

Hence, the inverse function is:

h^(-1)(x) = 3 ( (2e^x + 3) / (e^x - 1) )

I have assumed natural logarithms (base e). If base 10 logarithms, just change e to 10

Sep 18, 2017

y = (9+ (6*10^x))/(10^x - 1)

Explanation:

To find the inverse, let us switch the x and y variables, denoting h(x) as y

y = log((x+9)/(x-6)) implies x = log((y+9)/(y-6))

Assuming log(x) = log_10 (x),

Adding a base 10 to each side of the equation to cancel out the log, since 10^log(x) = x

10^x = 10^(log((y+9)/(y-6)))

10^x = (y+9)/(y-6)

Multiplying both sides by y-6 and subtracting y

10^x(y-6) = y+9 implies 10^xy - (6*10^x)-y = 9

Taking all y terms to one side:

10^xy-y = 9 + (6*10^x)

y(10^x - 1) = 9+ (6*10^x)

y = (9+ (6*10^x))/(10^x - 1)

I am aware of the other variations in which this answer could be rewritten, but you can work off of this answer to your preference.
Hope this helped!