(c) exactly three real solutions 

Given, \(x^{\frac{3}{4}(\text{log}_2\,x)^2}\) \(+\text{log}_2\,x-\frac{5}{4}\) = √2

Taking log to the base 2 of both the sides, we have

\(\bigg[\frac{3}{4}^{(\text{log}_2\,x)^2+(\text{log}_2)-\frac{5}{4}}​​\bigg]\) log₂ x = log₂ √2 = log₂ \(2^{\frac{1}{2}}\) = \(\frac{1}{2}\)log₂ 2 = \(\frac{1}{2}\)

Let us assume log₂x = a. Then,

\(\bigg(\frac{3}{4}a^2+a-\frac{5}{4}\bigg)a\) = \(\frac{1}{2}\) ⇒ 3a³ + 4a² - 5a = 2

⇒ 3a³ + 4a² – 5a – 2 = 0. 

Using hit and trial method check for a = 1. 

f(a) = 3a³ + 4a² – 5a – 2 ⇒ f(1) = 3.1³ + 4.1² – 5.1 – 2 = 0 

∴ (a – 1) is a factor of 3a³ + 4a² – 5a – 2

∴ Now by dividing 3a³ + 4a² – 5a – 2 by (a – 1), we get 

3a³ + 4a² – 5a – 2 = (a – 1) (3a + 1) (a + 2) = 0

⇒ a = 1 or a = \(-\frac{1}{3}\) or a = - 2

⇒ log₂x = 1 or log₂x = \(-\frac{1}{3}\) or log₂x = - 2

⇒ x = 2¹ = 2 or x = 2^-1/3 or x = 2^-2 = \(\frac{1}{4}\)

∴ The given equation has exactly three real solutions, wherein x = 2^–1/3 is irrational.