Question 399942
wrtie the expression as one logarithm: 2logxy^1/3 + logxy^2 - 3logxy
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I think you mean base x:
{{{2log(x,y^(1/3)) + log(x,y^2) - 3log(x,y)}}}
= {{{log(x,y^(2/3)) + log(x,y^2) - log(x,y^3)}}}
= {{{log(x,y^(8/3)) - log(x,y^3)}}}
= {{{log(x,y^(-1/3))}}}
or = {{{(-1/3)log(x,y))}}}