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How do you evaluate #log_(1/3) (1/81)#?

1 Answer

Sep 12, 2016

#4#

Explanation:

recall that #log_2(8)=3, => 2^3=8#

Let #log_(1/3)(1/81)=a#

#(1/3)^a=1/81#

#(1/(3^a)) =1/81#

#3^a=81#

#a=4#

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