The correct option is D two rational values and one irrational value
Clearly, x>0
x{34(log3x)2+(log3x)−54}=√3
⇒34(log3x)2+(log3x)−54=logx√3
⇒34(log3x)2+(log3x)−54=12log3x
⇒3(log3x)3+4(log3x)2−5(log3x)−2=0, log3x≠0
Let log3x=t
Then 3t3+4t2−5t−2=0, t≠0
Clearly t=1 is a solution.
So, (t−1)(3t2+7t+2)=0
⇒(t−1)(t+2)(3t+1)=0
⇒t=1,−2,−13
∴log3x=1,−2,−13
⇒x=3,19,13√3
Clearly, x takes two rational values and one irrational value.