Question

If for $x,y\in R,x>0,y={\log }_{10}x+{\log }_{10}{x}^{1/3}+{\log }_{10}{x}^{1/9}+\dots \text{upto}\infty$ terms and $\frac{2+4+6+\dots +2y}{3+6+9+\dots +3y}=\frac{4}{{\log }_{10}x}$, then the ordered pair $(x,y)$ is equal to:

• A. $({10}^6,9)$
• B. $({10}^6,6)$
• C. $({10}^4,6)$
• D. $({10}^2,3)$

Answer 1

The correct option is A $({10}^6,9)$

$y={\log }_{10}x+{\log }_{10}{x}^{1/3}+{\log }_{10}{x}^{1/9}+\dots \infty$
$={\log }_{10}(x\cdot x^{1/3}\cdot x^{1/9}\dots \infty )$
$={\log }_{10}\left(x^{1+\frac{1}{3}+\frac{1}{9}+\dots \infty }\right)$
$={\log }_{10}\left(x^{\frac{1}{1-\frac{1}{3}}}\right)$
$={\log }_{10}{x}^{3/2}$
$\therefore y=\frac{3}{2}{\log }_{10}x$
Now,
$\frac{2+4+6+...+2y}{3+6+9+\dots +3y}=\frac{4}{{\log }_{10}x}$
$\Rightarrow \frac{2(1+2+3+\dots +y)}{3(1+2+3+\dots +y)}=\frac{4}{{\log }_{10}x}$
$\Rightarrow \frac{2}{3}=\frac{4}{{\log }_{10}x}$
$\Rightarrow {\log }_{10}x=6$
$\Rightarrow x={10}^6$
$\therefore y=\frac{3}{2}\times 6=9$