Example 1 If \(a^{x}=b^{y}=1994^{x}\) (where \(a, b\) are natural numbers), and \(\frac{1}{x}+\frac{1}{y}=\frac{1}{z}\). Then all possible values of \(2a+b\) are ( ).
(A) 1001
(B) 1001,3989
(C) 1001,1996
(D) \(1001,1996,3989\)
(1994, National Junior High School Mathematics League)
Given: From the known information, we have $a=1994^{\frac{x}{x}}, b=1994^{\frac{x}{y}}$. Therefore, $a b=1994^{\frac{x}{x}+\frac{z}{y}}=1994$ $=2 \times 997$ (prime factorization).
From the conditions, we know $a \neq 1, b \neq 1$, so, $a=2, b=997$ or $a=997, b=2$.
Thus, $2 a+b$ is 1001 or 1996.