Solve the Following Questions


If for $x, y \in \mathbf{R}, x>0$

$y=\log _{10} x+\log _{10} x^{1 / 3}+\log _{10} x^{1 / 9}+\ldots .$ upto $\infty$ terms and $\frac{2+4+6+\ldots+2 y}{3+6+9+\ldots+3 y}=\frac{4}{\log _{10} x}$, then the ordered pair $(\mathrm{x}, \mathrm{y})$ is equal to :

  1. $\left(10^{6}, 6\right)$

  2. $\left(10^{4}, 6\right)$

  3. $\left(10^{2}, 3\right)$

  4. $\left(10^{6}, 9\right)$

Correct Option: , 4


$\frac{2(1+2+3+\ldots+y)}{3(1+2+3+\ldots+y)}=\frac{4}{\log _{10} x}$

$\Rightarrow \log _{10} x=6 \Rightarrow x=10^{6}$


$y=\left(\log _{10} x\right)+\left(\log _{10} x^{\frac{1}{3}}\right)+\left(\log _{10} x^{\frac{1}{9}}\right)+. . \infty$

$=\left(1+\frac{1}{3}+\frac{1}{9}+\ldots \infty\right) \log _{10} x$

$=\left(\frac{1}{1-\frac{1}{3}}\right) \log _{10} x=9$

So, $(x, y)=\left(10^{6}, 9\right)$

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