Question

If $x^4+y^4=47x^2{y}^2$ then show that $\log \left(\frac{x^2+y^2}{7}\right)=\log x+\log y$

Answer 1

$x^4+y^4=47x^2{y}^2$

$\Rightarrow {(x^2+y^2)}^2-2x^2{y}^2=47x^2{y}^2$

$\Rightarrow {(x^2+y^2)}^2=47x^2{y}^2+2x^2{y}^2$

$\Rightarrow {(x^2+y^2)}^2=49x^2{y}^2$

$\Rightarrow x^2+y^2=7xy$

$\Rightarrow \frac{x^2+y^2}{7}=xy$

Taking $\log$ to both sides, we get

$\Rightarrow \log \left(\frac{x^2+y^2}{7}\right)=\log \left(xy\right)$

$\therefore \log \left(\frac{x^2+y^2}{7}\right)=\log x+\log y$

Hence proved.