How do you condense $2 {log}_{3} x - 3 {log}_{3} y + {log}_{3} 8$ $2log_3 x -3log_3 y +log_3 8$ ?

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Answer 1 · 2016-06-15T04:58:18

$2 {log}_{3} x - 3 {log}_{3} y + {log}_{3} 8 = {log}_{3} (\frac{8 x^{2}}{y^{3}})$ $2log_(3)x-3log_(3)y+log_(3)8=log_3((8x^2)/y^3)$

Using $p {log}_{a} m = {log}_{a} m^{p}$ $plog_am=log_am^p$ and $log a + log b - log c = log (\frac{a b}{c})$ $loga+logb-logc=log((ab)/c)$

$2 {log}_{3} x - 3 {log}_{3} y + {log}_{3} 8$ $2log_(3)x-3log_(3)y+log_(3)8$

= ${log}_{3} x^{2} - {log}_{3} y^{3} + {log}_{3} 8$ $log_(3)x^2-log_(3)y^3+log_(3)8$

= ${log}_{3} (\frac{8 x^{2}}{y^{3}})$ $log_3((8x^2)/y^3)$

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