Question

If $x^{\frac{3}{4({\log }_{3}x{)}^2}+{\log }_{3}x-\frac{5}{4}}=\sqrt{3}$, then $x$ has

• A. one positive integral value
• B. one irrational value
• C. two positive rational values
• D. None of these

Answer 1

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

one positive integral value
one irrational value
two positive rational values

$x^{\frac{3}{4({\log }_{3}x{)}^2}+{\log }_{3}x-\frac{5}{4}}=\sqrt{3}$
Taking log on both sides on base 3, we have
$(\frac{3}{4({\log }_{3}x{)}^2}+{\log }_{3}x-\frac{5}{4})\times {\log }_{3}x={\log }_3\sqrt{3}[\because \log x^m=m\log x]$
$\Rightarrow (\frac{3}{4({\log }_{3}x{)}^2}+{\log }_{3}x-\frac{5}{4}){\log }_{3}x=\frac{1}{2}$
Let ${\log }_{3}x$ be $y$.
$\Rightarrow (\frac{3}{4y^2}+y-\frac{5}{4})y=\frac{1}{2}$
$\Rightarrow 4y^3-5y^2-2y+3=0$
$\Rightarrow (y-1)(3y^2+7y+2)=0$
$\Rightarrow y=1,\frac{-1}{3},-2$
$\Rightarrow x=3,3^{-1/3},\frac{1}{9}$
Hence, options A,B,C are correct.