If ${\log }_{10}x+\frac{1}{2}{\log }_{10}x+\frac{1}{4}{\log }_{10}x+\cdots =y$  and $\frac{1+3+5+\cdots +\left(2y-1\right)}{4+7+10+\cdots +\left(3y+1\right)}=\frac{20}{7{\log }_{10}x}\forall x,y\in N$ , then

$1+\frac{1}{2}+\frac{1}{4}+\cdots$  sum of $\infty$  terms in G.P

$=\left({\log }_{10}x\right)\left[\frac{1}{1-\frac{1}{2}}\right]$

$=2.{\log }_{10}x$

$\frac{1+3+5+\cdots +\left(2y-1\right)}{4+7+10+\cdots +\left(3y+1\right)}=\frac{20}{7{\log }_{10}x}$

$\frac{\frac{y}{2}\left[1+\left(2y-1\right)\right]}{\frac{y}{2}\left[4+\left(3y+1\right)\right]}=\frac{20}{7\left(\frac{y}{2}\right)}$

$\frac{2y}{3y+5}=\frac{40}{7y}$

$\frac{y}{3y+5}=\frac{20}{7y}$

$\Rightarrow 7y^2=60y+100$

$\Rightarrow 7y^2-60y-100=0$

$\Rightarrow 7y^2-70y+10y-100=0$

$7y\left(y-10\right)+10\left(y-10\right)=0$

$\left(y-10\right)\left(7y+10\right)=0$

$\therefore y=10$

$2.{\log }_{10}x=y$

${\log }_{10}{x}^2=10$

$x^2={10}^{10}$

$\Rightarrow x=\sqrt{{10}^{10}}={10}^5$

$\left(1\right){\log }_{y}x={\log }_{10}{10}^5=5$

$\left(3\right){\log }_y{x}^2={\log }_{10}{10}^{10}=10$

$\left(4\right){\log }_{y^5}x=\frac{1}{5}{\log }_{y}x=\frac{1}{5}\left(5\right)=1$