Question

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

Answer 1

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

`y=(1/2)^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`