How do you evaluate #log_4 1/64#?

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Answer 1 · 2015-07-09T14:15:39

#log_b 1=0#, for any base.

So we get #1/64#

Answer 2 · 2015-07-09T14:51:15

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

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#