(d) 3Taking log of both the sides to base 3, we have,\(\big[(log_3\,x)^2-rac{9}{2}log_3\,x+5\big]\) log3x = log333/2 = \(rac{3}{2}\) (∵ log33 = 1)⇒ 2(log3x)3 – 9(log3x)2 + 10 log3x – 3 = 0 ⇒ 2y3 – 9y2 + 10y – 3 = 0 (Take log3x = y)⇒ (y – 1) (y – 3) (2y – 1) = 0 (Factorising) ⇒ (log3x– 1) (log3x – 3) (2 log3x – 1) = 0 ⇒ log3x = 1, log3x = 3, 2 log3x = 1 ⇒ x = 31, x = 33, x2 = 31⇒ \(x\) = (3, 27, √3)∴ There are three solutions.