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!