Question

How do you evaluate `log_343 (1/7)`?

Answer 1

`-1/3`

Explanation:

Let `n = log_343 (1/7)`

`343^n=1/7`

Rewritten using all bases of 7:
`(7^3)^n=7^(-1)`

`7^(3n)=7^(-1)`

`3n=-1`

`n=-1/3`