Log question

I assume you have this :

2^(-log₂ 5)   if so.....we can write this as

2^(log₂ 5^-1)  =

2^[log₂ (1/5) ]  =

1/5

This goes back to the property that says  that  ...   b^(log_b a)  = a      where b is a positive base

Go for it CP, logs are not my thing.

Look at this, gibson338...

-log₂ 5  =  (-1) log₂ 5      and by a property of logs, we have

a log b   =  log b^a     ...... so......

(-1)log₂ 5     becomes   log₂ 5^-1     and 5^-1   = 1/5

So   we end up with     2^ [log₂ 1/5]