How do I get from 2^(x-log2 x) to 2^x/x

Question

How do I get from 2^(x-log2 x) to 2^x/x

Answer 1

2^x-log₂x

 

Log identity #1: a^b-c = a^b/a^c.  Hence:

 

2^x-log₂x = 2^x/2^log₂x

 

Log identity #2: a^log_ab = b.  Hence 2^log₂x = x:

 

2^x-log₂x = 2x/2^log₂x = 2^x/x

Answer 2

First, you can separate your original log into two pieces that are divided:

 

2^x/(2^log₂x)  (This is a common rule for combining logs of the same base)

 

Now, you need to recognize that anything raised to it's own base log like that, is just the "x" part of that equation.  Hard to explain in words, but maybe this makes sense:

 

(something)^(log _something x) = x

 

Basically, you are reversing the log process and getting right back to just x.  It's the foundational definition of what a log is.

 

This allows you to simply your equation to just 2^x/x