How do you condense $2logx-(3logy+logz)$ ?

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Answer 1 · 2016-12-03T19:15:33

The condensed answer is $log((x^2)/(y^3z))$ .

$2logx-(3logy+logz)$ .

$=logx^2-(logy^3+logz)$ .

$=logx^2-logy^3z$ .

$=log((x^2)/(y^3z))$ . (Answer).

How do you condense 2logx-(3logy+logz)2logx-(3logy+logz)?