# What is `log _9 (1/3)` ?

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-1/2

Let `log_9(1/3)=x`

Since `log_ax=n` means that `a^n=x` we can write:

`9^x=1/3`

`rArr (3^2)^x=1/3`

`rArr 3^(2x)=1/3`

Taking the log to the base 3 on both sides:

`log_3(3^(2x))=log_3(1/3)`

`rArr 2x=-1`

`rArr x=-1/2`

**Therefore**, `log_9(1/3)=-1/2.`

**Sources:**