\(9^{y} - 4 \times 3^{y} + 3 = 0\) \(\equiv (3^{2})^{y} - 4 \times 3^{y} + 3 = 0\) \((3^{y})^{2} - 4 \times 3^{y} + 3 = 0\) Let \(3^{y}\) be r. Then, \(r^{2} - 4r + 3 = 0\) Solving the equation, \(r^{2} - 3r - r + 3 = 0\) \(r(r - 3) - 1(r - 3) = 0\) \((r - 3)(r - 1) = 0\) \(\therefore \text{r = 3 or 1}\) Recall, \(3^{y} = r\) \(3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}\) \(\implies \text{y = 1 or 0}\)