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How do you condense #2log_3 x -3log_3 y +log_3 8#?

1 Answer

Shwetank Mauria

Jun 15, 2016

#2log_(3)x-3log_(3)y+log_(3)8=log_3((8x^2)/y^3)#

Explanation:

Using #plog_am=log_am^p# and #loga+logb-logc=log((ab)/c)#

#2log_(3)x-3log_(3)y+log_(3)8#

= #log_(3)x^2-log_(3)y^3+log_(3)8#

= #log_3((8x^2)/y^3)#

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