Question

The equation $\frac{3}{4}{\left({\log }_{2}x\right)}^2+{\log }_{2}x\left(-\frac{5}{4}\right)={\log }_2\sqrt{2}$ has

• A. at least one real solution
• B. exactly three real solutions
• C. exactly one irrational solution
• D. complex roots

Answer 1

The correct option is C

exactly one irrational solution

Explanation for the correct option:

Finding the roots of the equation:

Step 1: Substitute ${\log }_{2}x=t$

Given,

$\frac{3}{4}{\left({\log }_{2}x\right)}^2+{\log }_{2}x\left(-\frac{5}{4}\right)={\log }_2\sqrt{2}$

Substitute ${\log }_{2}x=t$

$$\begin{gathered} \Rightarrow \left(\frac{3}{4}\right)t^2+t\left(-\frac{5}{4}\right)=\frac{1}{2} \\ \Rightarrow 3t^2-5t=2 \\ \Rightarrow 3t^2-5t-2=0 \\ \Rightarrow 3t^2-6t+t-2=0 \\ \Rightarrow 3t\left(t-2\right)+\left(t-2\right)=0 \\ \Rightarrow \left(3t+1\right)\left(t-2\right)=0 \end{gathered}$$

we get,

$$\begin{gathered} t-2=0 \\ \Rightarrow t=2 \\ or3t+1=0 \\ \Rightarrow t=-\frac{1}{3} \end{gathered}$$

Therefore,

$t=2,–\frac{1}{3}$

Step 2: Finding the value of $x$

Since, ${\log }_{2}x=t$, as substituted above,

${\log }_{2}x=2,–1/3$

Therefore, we get,

$$\begin{gathered} {\log }_2\left(x\right)=-\frac{1}{3} \\ \Rightarrow x=\frac{1}{2^{{\displaystyle \frac{1}{3}}}} \end{gathered}$$

$$\begin{gathered} {\log }_2\left(x\right)=2 \\ \Rightarrow x=4 \end{gathered}$$

So, two solutions $4,\frac{1}{2^{{\displaystyle \frac{1}{3}}}}$ are real

Hence, the correct answer is option (C)