Question 105931
{{{logx^2+logy-3logx-logy^2}}}
:
Adding logs means multiply so can have:
{{{log((x^2y)) - 3log((x)) - log((y^2))}}}
:
The exponent equiv of 3*log
{{{log((x^2y)) - log((x^3)) - log((y^2))}}}
:
We have to combine the minus logs like this:
{{{log((x^2y)) - log((x^3y^2))}}}
:
Subtract means divide
{{{log((x^2y/x^3y^2))}}} = {{{log((1/xy))}}}
:
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