$$
\begin{array}{l}
2 \log _{10}(x-2 y)=\log _{10} x+\log _{10} y \\
(x-2 y)^{2}=x y \\
x^{2}-5 x y+4 y^{2}=0 \\
(x-y)(x-4 y)=0 \\
d=\frac{x}{y}=1 \text { or } 4
\end{array}
$$

Check: When $x=y$, L.H.S. $=2 \log _{10}(y-2 y)=2 \log _{10}(-y)$, R.H.S. $=\log _{10} y+\log _{10} y$
When $y>0$. L.H.S. is undefined and R.H.S. is well defined, rejected
When $y=0$, L.H.S. is undefined and R.H.S. is undefined, rejected
When $y0$. L.H.S. is well defined and R.H.S. is well defined, accepted
$$
\therefore d=4 \text { only }
$$
I3.4 若實數 $x$ 及 $y$ 滿足方程 $2 \log _{10}(x+2 c y)=\log _{10} x+\log _{10} y$, 求 $d=\frac{x}{y}$ 的值。
If real numbers $x$ and $y$ satisfy the equation $2 \log _{10}(x+2 c y)=\log _{10} x+\log _{10} y$, determine the value of $d=\frac{x}{y}$.