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

1 Answer
Apr 18, 2016

Answer:

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

Explanation:

#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!