How do you evaluate #log_64 8#?

2 Answers
Sep 18, 2016

#log_64 8=1/2#.

Explanation:

We will use the Defn. of #log# fun. to find the desired value.

Recall that, #log_ba=x iff b^x=a#.

So, #log_64 8=x rArr 64^x=8 rArr (8^2)^x=8 rArr 8^(2x)=8^1#.

As #log# fun. is #1-1#, we must have, #2x=1," giving, "x=1/2#.

#:. log_64 8=1/2#.

Sep 18, 2016

#log_(64)8=1/2#

Explanation:

When I write #log_ab=c#, I ask to what power I raise the base #a# to get #b#; the answer here is #c#, i.e. #a^c=b#. For #log_(10)10#, #log_(10)100#, #log_(10)1000#, #log_(10)1000000#, I can immediately write #1,2,3,6# without thinking too much.

For your problem, #log_(64)8#, I look for the power to which I raise #64# to get #8#. But of course, #8^2=64#, or #sqrt64=8#, #64^(1/2)=8#. So #log_(64)8=1/2#.