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

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Answer 1 · 2016-11-13T17:53:35

$-1/3$

Let $n = log_343 (1/7)$

$343^n=1/7$

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

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

$3n=-1$

$n=-1/3$

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