Question over logarithms?

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1 Answer
Dec 12, 2017

Let's start by showing that #f(g(x))=x#, which we do by plugging in #g(x)# into #f(x)#:
#f(g(x))=3+log_2(g(x)-1)=3+log_2(1+2^(x-3)-1)#

The #1#'s inside the logarithm cancel, so we're left with:
#3+cancel(log_2)(cancel(2)^(x-3))=x+cancel(3-3)=x#

And now for #g(f(x))#:
#g(f(x))=1+2^(f(x)-3)=1+2^(3+log_2(x-1)-3)#

This time the #3#'s in the exponent cancels, giving:
#1+cancel(2)^(cancel(log_2)(x-1))=x+cancel(1-1)=x#

Since #f(g(x))=x# and #g(f(x))=x#, we know that the two functions are inverses.