How do you write #9^(-1/2)=1/3# in Log form?

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Answer 1 · 2016-09-02T06:36:07

#log_9(1/3)=-1/2#

First let's remember that:

#x^b=y# is saying "how many times do we multiply x by itself to arrive at y"

and that in log form, it's

#log_xy=b# is saying "b is the number of times we x by itself in order to arrive at y"

So we can convert:

#9^(-1/2)=1/3# into

#log_9(1/3)=-1/2#