How do you evaluate #log_(2/5) (125/8)#?

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Answer 1 · 2016-11-24T02:18:06

#- 3#

Useful Hint for this question:

#log_aa^n=nlog_aa=n#

Solution :

#log_(2/5)(125/8)=log_(2/5)(5/2)^3#

= #log_(2/5)(2/5)^(-3)#

=#(-3)log_(2/5)(2/5)#

= -3.

[As #log _a a=1#]