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

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

How do you write 9^(-1/2)=1/39^(-1/2)=1/3 in Log form?