4. Given $41^{x}=2009,7^{y}=2009$. Then the value of $\frac{1}{x}+\frac{2}{y}$ is . $\qquad$
4. 1 .

From $41^{x}=2009$, we get $41^{x y}=2009^{y}$.
$$
\begin{array}{l}
\text { Also } 7^{y}=2009 \Rightarrow 7^{2 y}=2009^{2} \\
\Rightarrow 49^{y}=2009^{2} \Rightarrow 49^{x y}=2009^{2 x} .
\end{array}
$$
$$
\text { Also } 41 \times 49=2009 \Rightarrow 41^{x y} \times 49^{x y}=2009^{x y} \text {. }
$$

Therefore $2009^{y} \times 2009^{2 x}=2009^{x y}$
$$
\begin{array}{c}
\Rightarrow 2009^{2 x+y}=2009^{x y} \Rightarrow 2 x+y=x y \\
\Rightarrow \frac{2}{y}+\frac{1}{x}=1 .
\end{array}
$$