-1 and 0
-1 and 1
1 and 3
0 and 1
Correct answer is D
9y−4×3y+3=0
≡(32)y−4×3y+3=0
(3y)2−4×3y+3=0
Let 3y be r. Then,
r2−4r+3=0
Solving the equation,
r2−3r−r+3=0
r(r−3)−1(r−3)=0
(r−3)(r−1)=0
∴
Recall, 3^{y} = r
3^{y} = 3 = 3^{1} \text{ or } 3^{y} = 1 = 3^{0}
\implies \text{y = 1 or 0}
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