Question

If for $x,y\in R,x>0$,
$y={\log }_{10}x+{\log }_{10}{x}^{1/3}+{\log }_{10}{x}^{19}+\dots ..$ 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. $\left(10^6,6\right)$
• B. $\left(10^4,6\right)$
• C. $\left(10^2,3\right)$
• D. $\left(10^6,9\right)$

Answer 1

Answer: (D)

$y={\log }_{10}x+{\log }_{10}{x}^{1/3}+{\log }_{10}{x}^{1/9}+\dots \infty$
$={\log }_{10}\left(x\cdot x^{1/3}\cdot x^{1/9}\dots \infty \right)$
$={\log }_{10}\left(x^{1+\frac{1}{3}+\frac{1}{9}\dots \infty }\right)$
$={\log }_{10}\left(\frac{1}{1-\frac{1}{3}}\right)$
$={\log }_{10}{x}^{3/2})$
$\therefore y=\frac{3}{2}{\log }_{10}x$
Now,
$2+4+6+\dots +2y$
$\frac{2+6+9+\dots +3y}{3+6}=\frac{4}{{\log }_{10}x}$
$\Rightarrow \frac{2(1+2+3+\dots +y)}{3(1+2+3+\dots +y)}=\frac{4}{{\log }_{10}x}$
$\Rightarrow \frac{2}{3}=\frac{4}{{\log }_{10}x}$
$\Rightarrow {\log }_{10}x=6$
$\Rightarrow x=10^6$
$\therefore y=\frac{3}{2}x6=9$