How do you simplify # log_(5) (1/15)#?

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Answer 1 · 2016-06-12T01:15:11

#=log_5(1) - log_5(15)#

Consider that #log(1)# of anything is equal to #0#.

#=0 - log_5(15)#

#=-log_5(5 xx 3)#

Recall that #log(a xx n) = loga + logn#

#= -(log_5(5) + log_5(3))#

Finally, remember that #log_a(a) = 1#.

#= -log_5(3) - 1#

Hopefully this helps!