What is the inverse of #f(x)= -log_3 (x^3)-3log_3(x-3)# ?

1 Answer
George C.
Mar 12, 2016

#f^(-1)(y) = sqrt(3^(-y/3)+9/4)+3/2#

Explanation:

Assuming we are dealing with #log_3# as a Real valued function and inverse of #3^x#, then the domain of #f(x)# is #(3, oo)#, since we require #x > 3# in order that #log_3(x-3)# be defined.

Let #y = f(x)#

#= -log_3(x^3)-3log_3(x-3)#

#=-3 log_3(x)-3 log_3(x-3)#

#=-3 (log_3(x)+log_3(x-3))#

#=-3 log_3(x(x-3))#

#=-3 log_3(x^2-3x)#

#=-3 log_3((x-3/2)^2-9/4)#

Then:

#-y/3 = log_3((x-3/2)^2-9/4)#

So:

#3^(-y/3) = (x-3/2)^2-9/4#

So:

#3^(-y/3)+9/4 = (x-3/2)^2#

So:

#x-3/2 = +-sqrt(3^(-y/3)+9/4)#

In fact, it must be the positive square root since:

#x-3/2 > 3-3/2 > 0#

So:

#x = sqrt(3^(-y/3)+9/4)+3/2#

Hence:

#f^(-1)(y) = sqrt(3^(-y/3)+9/4)+3/2#

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