How do you evaluate #log_4 (1/2)#?
2 Answers
Jan 7, 2016
Explanation:
Using the following properties:

#log_a(a) = 1# 
#log(a^x) = xlog(a)#
we have
Jan 7, 2016
Explanation:
Based on the meaning of
If
then
We know that
So
Also
Which leads us to asking for what value of
Since
and