How do you find the inverse of y=log_(1/2) xy=log12x?

2 Answers
Jan 9, 2016

It's (1/2)^x(12)x

log_(1/2)(x) = ln(x)/ln(1/2)log12(x)=ln(x)ln(12)

by logic you put (1/2)^x(12)x

ln((1/2)^x)/ln(1/2)ln((12)x)ln(12) with the identity aln(b) = ln(b^a)aln(b)=ln(ba) you transform

xln(1/2)/ln(1/2) = xxln(12)ln(12)=x

You find the identity application so y = (1/2)^xy=(12)x is the inverse function because

f@f^(-1) = xff1=x

Jan 9, 2016

y=(1/2)^xy=(12)x

Explanation:

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