Question

If $x^2+y^2=47xy$ then show that $\log \left(\frac{x+y}{7}\right)=\frac{1}{2}(\log x+\log y).$

Answer 1

Given that,

$x^2+y^2=47xy$

Add $2xy$ both sides,

$\begin{array}{c}{x}^2+y^2+2xy=47xy+2xy\\ {\left(x+y\right)}^2=49xy\end{array}$

Square root both sides,

$x+y=7\sqrt{xy}$

Substitute the above equation in,

$\begin{array}{c}\log \left(\frac{x+y}{7}\right)=\frac{1}{2}\left(\log x+\log y\right)\\ \log \left(\frac{7\sqrt{xy}}{7}\right)=\frac{1}{2}\left(\log x+\log y\right)\\ \log \left(\frac{7{\left(xy\right)}^{\frac{1}{2}}}{7}\right)=\frac{1}{2}\left(\log x+\log y\right)\\ \log \left({\left(xy\right)}^{\frac{1}{2}}\right)=\frac{1}{2}\left(\log x+\log y\right)\\ \frac{1}{2}\left(\log \left(xy\right)\right)=\frac{1}{2}\left(\log x+\log y\right)\\ \frac{1}{2}\left(\log x+\log y\right)=\frac{1}{2}\left(\log x+\log y\right)\end{array}$

LHS=RHS