Question

If $\log \left(\frac{x+y}{3}\right)=\frac{1}{2}\log x+\frac{1}{2}\log y$ then show that, $\frac{x}{y}+\frac{y}{x}=7$
OR
Show that, ${\log }_b{a}^5.{\log }_c{b}^3.{\log }_a{c}^7=105$

Answer 1

$\left(1\right)\log \left(\frac{x+y}{3}\right)=\frac{1}{2}\log x+\frac{1}{2}\log y$

$=\log x^{1/2}+\log y^{1/2}$

$=\log (xy{)}^{1/2}$

$\therefore \left(\frac{x+y}{3}\right)=(xy{)}^{1/2}$

$\therefore \frac{x^2+2xy+y^2}{9}=xy$

$\therefore x^2+y^2=9xy-2xy$

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

$\therefore \frac{x}{y}+\frac{x}{y}=7$

$\left(2\right){\log }_b{a}^5.{\log }_c{b}^3.{\log }_a{c}^7=105$

${\log }_n{n}^x=x{\log }_{n}m$

$=n\frac{\log m}{\log n}$

$\therefore$ LHS$=5\frac{\log a}{\log b}\times 3\frac{\log b}{\log c}\times 7\frac{\log c}{\log a}$

$=5\times 3\times 7$

$=105$