How do you find the inverse of #y=log_(1/2) x#?

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Answer 1 · 2016-01-09T19:13:26

It's #(1/2)^x#

#log_(1/2)(x) = ln(x)/ln(1/2)#

by logic you put #(1/2)^x#

#ln((1/2)^x)/ln(1/2)# with the identity #aln(b) = ln(b^a)# you transform

#xln(1/2)/ln(1/2) = x#

You find the identity application so #y = (1/2)^x# is the inverse function because

#f@f^(-1) = x#

Answer 2 · 2016-01-09T20:55:27

#y=(1/2)^x#

Given #color(white)(....)y=log_(1/2)(x) #

By the nature of what a log is, another way of writing the given relationship is:

# (1/2)^y=x#

As it turns out this is exactly the format needed for the inverse function. All you have to now do is swap the variable letters round to give:

So if #f(x)=log_(1/2)(x)#

Then #f^(-)(x) =(1/2)^x #

Or, if you prefer #-> y=(1/2)^x#