How do you simplify $log_5 (1/15)$ ?

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Answer 1 · 2016-05-31T04:58:44

$-log 15/log 5=-1.6826$

I am following Phoenix Flare's answer, for approximation purpose.

The answer given is

$log_5 1- log_5 15$

$=0-log 15/log 5=-1.6826$ , nearly,, using

$log_b 1 = 0 and log_b a = log_c a/log_c b$ ,

where c is at your choice.

For that matter, the answer could also be given as

$-ln 15/ln 5= -1.6826$ , nearly.

How do you simplify log_5 (1/15) log_5 (1/15)?