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.