Question

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

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

Answer 1

Answer: (A), (B), (C)

The correct options are
A at least one real solution
B exactly three real solutions
C exactly one irrational solution

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

Take log both sides on base $x$ and use ${\log }_{x}x=1$
$\Rightarrow \frac{3}{4}{\left({\log }_{2}x\right)}^2+{\log }_{2}x-\frac{5}{4}={\log }_x\sqrt{2}$

$\Rightarrow \frac{3}{4}{\left({\log }_{2}x\right)}^2+{\log }_{2}x-\frac{5}{4}=\frac{1}{2{\log }_{2}x},$

$[\because \log a^x=x\log a\text{\&}{\log }_{b}a=\frac{1}{{\log }_{a}b}]\Rightarrow 3{\left({\log }_{2}x\right)}^3+4{\left({\log }_{2}x\right)}^2-5\left({\log }_{2}x\right)-2=0$

Put ${\log }_{2}x=y$
$\therefore {3y}^3+{4y}^2-5y-2=0\Rightarrow (y-1)(y+2)(3y+1)=0$
$\Rightarrow y=1,-2,\frac{-1}{3}$
$\Rightarrow {\log }_{2}x=1,-2,\frac{-1}{3}$
$\Rightarrow x=2,\frac{1}{2^{\frac{1}{3}}},\frac{1}{4}$