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

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Answer 1 · 2016-06-15T04:58:18

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

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

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

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

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

How do you condense 2log_3 x -3log_3 y +log_3 82log_3 x -3log_3 y +log_3 8?