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

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1 Answer

Aug 23, 2016

3

Explanation:

#log_(1/5) (1/125) = log_(1/5) (1/5^3) = log_(1/5) (1/5)^3#

#= 3log_(1/5) (1/5)#

#=3#

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