Question 974318
<pre>
If {{{x^2 + y^2 = 7xy}}} then {{{ log ((x + y)) + log (1/3) }}}  = ?
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I don't know what convoluted stuff that other person who responded wrote, but all that MUMBO JUMBO
seems, in my opinion, to have absolutley nothing to do with the expression to be evaluated!

If {{{x^2 + y^2 = 7xy}}} then {{{log ((x + y)) + log (1/3)}}}  = ?
   {{{(x + y)^2 = x^2 + 2xy + y^2}}}
   {{{(x + y)^2 = x^2 + y^2 + 2xy}}}
   {{{(x + y)^2 = 7xy + 2xy}}} ---- Substituting 7xy for {{{x^2 + y^2}}}
   {{{(x + y)^2 = 9xy}}}
{{{sqrt((x + y)^2) = 0+- sqrt(9xy)}}} ----- Taking the square root of each side
     {{{x + y = 0 +- 3sqrt(xy)}}}

{{{highlight(log ((x + y)) + log (1/3))}}} = ?
{{{log ((3sqrt(xy))) + log ((1/3))}}} --- Substituting ONLY the POSITIVE value ({{{3sqrt(xy)}}}), for x + y
{{{log (((3sqrt(xy))(1/3)))}}} ----- Applying {{{log (b, (c)) + log (b, (d))}}} = {{{log (b, (c*d))}}}
{{{log ((cross(3)sqrt(xy))(1/cross(3)))}}} = {{{highlight(log (sqrt(xy)))}}}, or {{{highlight(log ((xy)^(1/2)))}}}</pre>