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

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Answer 1 · 2016-04-08T17:31:01

#x=-1#

#log_(1/125) 125=x#

#125=(1/125)^x#

#5^3=(5^-3)^x#

#5^3=5^(-3x)#

#"so equation has the same base;"#

#3=-3x#

#x=-1#