How do you calculate $log_9(1/7)$ ?

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Answer 1 · 2016-05-30T16:33:24

$-loh 7/log 9=-0.8856$ , nearly.

Use $log_b a^n=n log_b a and log_b a=log_c a/log_c b$

Here, $log_9(1/7)=log_9(7^(-1))=-log_9 7= - log 7/log 9$

How do you calculate log_9(1/7)log_9(1/7)?