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!