The equation ${x^{\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}} = \sqrt 2 $ has
A. at least one real solution
B. exactly three real solutions
C. exactly one irrational solution
D. complex roots

Hint: Simplify the given expression by taking log both sides. Then, substitute ${\log _2}x = t$ and find the factor of the equation by hit and trial method. Then, write all the factors of the equation. Next, equate each of the factors to 0 to find the value of $t$ and hence the value of $x$.

Complete step-by-step answer:
We will solve the equation ${x^{\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}} = \sqrt 2 $
Let us simplify the equation by taking log with base 2 on both sides.
${\log _2}\left( {{x^{\dfrac{3}{4}{{\left( {\log x} \right)}^2} + {{\log }_2}x - \dfrac{5}{4}}}} \right) = {\log _2}\sqrt 2 $
But, we know that ${\log _a}{b^n} = n{\log _a}b$
$\left( {\dfrac{3}{4}{{\left( {{{\log }_2}x} \right)}^2} + {{\log }_2}x - \left( {\dfrac{5}{4}} \right)} \right){\log _2}x = {\log _2}\sqrt 2 $
Let ${\log _2}x = t$
$\left( {\dfrac{3}{4}{t^2} + t - \dfrac{5}{4}} \right)t = \dfrac{1}{2}$
On cross-multiplying and simplifying the equation , we will get,
$
  \dfrac{3}{4}{t^3} + {t^2} - \dfrac{5}{4}t = \dfrac{1}{2} \\
   \Rightarrow \dfrac{3}{2}{t^3} + 2{t^2} - \dfrac{5}{2}t = 1 \\
   \Rightarrow 3{t^3} + 4{t^2} - 5t - 2 = 0 \\
$
We will use hit and trial to find the root of the above equation,
Put $t = 1$ in the above equation,
$
  3{\left( 1 \right)^3} + 4{\left( 1 \right)^2} - 5\left( 1 \right) - 2 \\
   = 3 + 4 - 5 - 2 \\
   = 7 - 7 \\
   = 0 \\
$
Hence, $t - 1$ is the factor of the equation, $3{t^3} + 4{t^2} - 5t - 2 = 0$
Therefore, we can divide the equation by $t - 1$ and write the equation as
$\left( {t - 1} \right)\left( {3{t^2} + 7t + 2} \right) = 0$
Factorise the above expression,
$
  \left( {t - 1} \right)\left( {3{t^2} + 6t + t + 2} \right) = 0 \\
   \Rightarrow \left( {t - 1} \right)\left( {3t\left( {t + 2} \right) + 1\left( {t + 2} \right)} \right) = 0 \\
   \Rightarrow \left( {t - 1} \right)\left( {3t + 1} \right)\left( {t + 2} \right) = 0 \\
$
Equate each factor to 0 and the value of $t$
$
  t - 1 = 0 \\
   \Rightarrow t = 1 \\
$
$
  3t + 1 = 0 \\
   \Rightarrow t = - \dfrac{1}{3} \\
$
And
$
  t + 2 = 0 \\
   \Rightarrow t = - 2 \\
$
Substitute back $t = {\log _2}x$ and hence find the value of $x$
$
  t = 1 \\
   \Rightarrow {\log _2}x = 1 \\
   \Rightarrow x = {2^1} \\
   \Rightarrow x = 2 \\
$
Similarly,
$
  t = - 2 \\
   \Rightarrow {\log _2}x = - 2 \\
   \Rightarrow x = {2^{ - 2}} \\
   \Rightarrow x = \dfrac{1}{4} \\
$
And
$
  t = - \dfrac{1}{3} \\
   \Rightarrow {\log _2}x = - \dfrac{1}{3} \\
   \Rightarrow x = {2^{ - \dfrac{1}{3}}} \\
$
Since, $x$ has three real values.
Therefore, there are exactly three real solutions for the given equation.
Hence, option C is correct.

Note: If the number is written of the form ${\log _a}b = x$, then $a$ is the base of the logarithmic function and $b = {a^x}$. Also, students must know the properties of logarithmic function, like ${\log _a}{b^n} = n{\log _a}b$ and ${\log _a}{a^n} = n$