How do you evaluate #log_(1/5) 25#?

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Answer 1 · 2016-08-21T18:16:56

#log_(1/5)25=-2#

Let #log_(1/5)25=x#, then

#(1/5)^x=25# or

#1/5^x=5^2# or

#5^(-x)=5^2#

i.e. #-x=2# or

#x=-2#

Hence #log_(1/5)25=-2#.