How do you calculate #log_(1/5) 125#?

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Answer 1 · 2016-04-18T01:59:58

First, use the change of base rule #log_an = logn/loga#

#log_(1/5)(125)#

#->log125/log(1/5)#

#->log(5^3)/log(5^-1)#

Now, use the following log rule: #logn^a = alogn#

#->(3log5)/(-1log5)#

#->-3#

Thus, #log_(1/5)(125) = -3#

Hopefully this helps!