Solve the following equations for $x\text{ and }y:$${\log }_{10}x+{\log }_{10}{x}^{1/2}+{\log }_{10}{x}^{1/4}+\dots =y$ and  $\frac{1+3+5+\dots +(2y-1)}{4+7+10+\dots +(3y+1)}=\frac{20}{7{\log }_{10}x}$

We have $y={\log }_{10}x+{\log }_{10}{x}^{1/2}+{\log }_{10}{x}^{1/4}+\dots$

$\begin{array}{l}\Rightarrow y=\left(1+\frac{1}{2}+\frac{1}{2^2}+\dots \right){\log }_{10}x\\ \Rightarrow y=\left(\frac{1}{1-\frac{1}{2}}\right){\log }_{10}x=2{\log }_{10}x\end{array}$

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

$\begin{array}{l}\Rightarrow \frac{y^2}{\frac{y}{2}(4+3y+1)}=\frac{20}{7{\log }_{10}x}\\ \Rightarrow \frac{2y}{(3y+5)}=\frac{20}{7{\log }_{10}x}\\ \Rightarrow \frac{y}{(3y+5)}=\frac{10}{7{\log }_{10}x}\\ \Rightarrow \frac{y}{(3y+5)}=\frac{10}{7y/2}=\frac{20}{7y}\\ \Rightarrow 7y^2-60y-100=0\\ \Rightarrow 7y^2-70y+10y-100=0\\ \Rightarrow 7y(y-10)+10(y-10)=0\\ \Rightarrow (y-10)(7y+10)=0\\ \Rightarrow y=10,-\frac{10}{7}\end{array}$

Thus, $2{\log }_{10}x=10,-\frac{10}{7}$

$\begin{array}{l}\Rightarrow {\log }_{10}x=5,-\frac{5}{7}\\ \Rightarrow x={10}^5,{10}^{-\frac{5}{7}}\end{array}$

Hence, the solutions are

$\Rightarrow x={10}^5,{10}^{-\frac{5}{7}};y=5,-\frac{5}{7}$