What is `log_125 (1/5)` ?

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Let `log_125(1/5)=x`

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

`125^x=1/5`

`rArr (5^3)^x=1/5`

`rArr 5^(3x)=1/5`

Taking the log to the base 5 on both sides:

`log_5(5^(3x))=log_5(1/5)`

`rArr 3x=-1`

`rArr x=-1/3`

**Therefore,** `log_125(1/5)=-1/3`

Here, change of base formula can also be applied to get the same result.

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