How do you condense $2 log x - 3 (log y + 2 log z)$ $2logx-3(logy+2logz)$ ?

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Answer 1 · 2016-06-05T04:41:24

$2 log x - 3 (log y + 2 log z) = log (\frac{x^{2}}{y^{3} z^{6}})$ $2logx−3(logy+2logz)=log(x^2/(y^3z^6))$

$2 log x - 3 (log y + 2 log z)$ $2logx−3(logy+2logz)$

= $2 log x - 3 log y - 6 log z$ $2logx−3logy-6logz$

= ${log x}^{2} - {log y}^{3} - {log z}^{6}$ $logx^2−logy^3-logz^6$

= $log (\frac{x^{2}}{y^{3} z^{6}})$ $log(x^2/(y^3z^6))$

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