How do you evaluate #log_4 1/64#?

2 Answers
Jul 9, 2015

#log_b 1=0#, for any base.

Explanation:

So we get #1/64#

Jul 9, 2015

If the question was intended to be: Find #log_4(1/64)#, then the answer is #-3#.

Explanation:

Remember that the log base 4 of a number is the exponent needed on 4 to get that number.

#1/64 = 1/4^3 = 4^(-3)#

Now what exponent do I need to put on 4, in order to get #4^(-3)#?

(Yes, it is a self-answering question. It's fun to think of other self-answering questions. OK, it might be nerdy fun.)

#log_4 4^(-3) = -3#