Question 815452
<pre>
if x*x+y*y=7xy, prove
log(x+y)=1/2(logx+logy)+log3

If x*x+y*y=7xy, prove log(x+y)=1/2(logx+logy)+log3
{{{x^2 + y^2 = 7xy}}}   Prove: {{{log ((x + y)) = (1/2)(log ((x)) + log ((y))) + log ((3))}}}
   {{{(x + y)^2 = x^2 + 2xy + y^2}}} ----- Squaring x + y
   {{{(x + y)^2 = x^2 + y^2 + 2xy}}}
   {{{(x + y)^2 = 7xy + 2xy}}} --- Substituting 7xy for {{{x^2 + y^2}}} (GIVEN)
   {{{(x + y)^2 = 9xy}}}
{{{sqrt((x + y)^2) = sqrt(9xy)}}} ------- Taking the square root of both sides
     {{{x + y = 3sqrt(xy)}}} <=== ONLY the POSITIVE square root on the right-side is needed here 

{{{log ((x + y)) = log ((3sqrt(xy)))}}} ----- Taking the log of both sides of the above equation
{{{log ((x + y)) = log ((3)) + log (sqrt((xy)))}}}
{{{log ((x + y)) = log ((3)) + log ((xy))^(1/2)}}}
{{{log ((x + y)) = log ((3)) + (1/2)log ((xy))}}}
{{{log ((x+ y)) = log ((3)) + (1/2)(log ((x)) + log ((y)))}}} <=== QED!</pre>