How do you solve for w in $logw=1/2logx + logy$ ?

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Answer 1 · 2016-05-18T21:33:54

$w = x^2y$

$0 = 1/2logx + logy - logw$

$0 = logx^2 + logy - logw$

$0 = log((x^2 xx y)/w)$

$10^0 = (x^2y)/w$

$1 = (x^2y)/w$

$w = x^2y$

Hopefully this helps!

How do you solve for w in logw=1/2logx + logylogw=1/2logx + logy?