Question

12. If log x-y/2 = (log x + log y), prove that x2 + y2 = 6xy.​

Answer 1

$Given:</p><p>log(x−y)/2=1/2(logx+logy)</p><p>Let us simplify</p><p>log(x−y)/2=1/2(logx×y)</p><p>log(x−y)/2=1/2logxy</p><p>log(x−y)/2=log(xy)1/2</p><p>By cross multiplying, we get</p><p>(x−y)2=4xy</p><p>x2+y2−2xy=4xy</p><p>x2+y2=4xy+2xy</p><p>x2+y2=6xy</p><p>Hence proved.</p><p></p><p>$

Answer 2

Answer:

logx + logy=logxy

so xy=(x-y) /2

x-y=2xy

(x-y) ^2=4x^2y^2

x^2+y^2-2xy=(2xy) ^2

x^2+y^2=2xy(2xy+1)