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How do you evaluate #log_225 15#?

1 Answer

Aug 28, 2016

#1/2#.

Explanation:

#x=log_(225) 15 rArr 225^x=15 rArr (15^2)^x=15rArr 15^(2x)=15^1#

#rArr 2x=1

#rArr x=1/2#

#rArr log_(225) 15=1/2#.

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