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How do you evaluate # Log _ 9(1/3)#?

1 Answer

Dec 15, 2015

#color(white)(xx)-1/2#

Explanation:

#color(white)(xx)log_9 (1/3)=log_(3^2) 3^-1#
#color(white)(xxxxxxxx)=1/2xx(-1)log_3 3#
#color(white)(xxxxxxxx)=-1/2#

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