If for x, y ∈ R, x>0, y= log_10 x+ log_10 x^1 / 3+ log_10 x^1 / 9+… upto ∞ terms and (2+4+6+…+2 y)/(3+6+9+…+3 y)=4/(log_10 x), then the ordered pair…

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