Equation involving square root and cube root . Olympiad Mathematics 

(1/3)lnx=(1/2)ln(3x) , (1/3)lnx=(1/2)ln(3)+(1/2)ln(3) , (1/2-1/3)lnx= - (1/2)ln3 , lnx= - 3ln3 , x=e^(-3ln3) , x=1/27 , test , (1/27)^(1/3)=(1/(3^3))^(1/3) , 1/((3^3)^(1/3))=1/3 , (3x)^(1/2)=(3/27)^(1/2) , (1/9)^(1/2)=1/3 , same , OK ,

0, 1/27.

Well done

X= 0.

X= 0, X=⅓.

Since both sides involves power of cube root and square root, we can take power of both sides (LCM of 2 and 3) to 6, we can straight get x² = (3x)³ = 27x³ and follows the rest.