Question

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

• A. all integral values
• B. two integral values and one irrational value
• C. all irrational values
• D. two rational values and one irrational value

Answer 1

The correct option is D two rational values and one irrational value

Clearly, $x>0$
$x^{\left\{\frac{3}{4}\right({\log }_{3}x{)}^2+\left({\log }_{3}x\right)-\frac{5}{4}\}}=\sqrt{3}$
$\Rightarrow \frac{3}{4}({\log }_{3}x{)}^2+({\log }_{3}x)-\frac{5}{4}={\log }_x\sqrt{3}$
$\Rightarrow \frac{3}{4}({\log }_{3}x{)}^2+({\log }_{3}x)-\frac{5}{4}=\frac{1}{2{\log }_{3}x}$
$\Rightarrow 3({\log }_{3}x{)}^3+4({\log }_{3}x{)}^2-5\left({\log }_{3}x\right)-2=0,{\log }_{3}x\ne 0$

Let ${\log }_{3}x=t$
Then $3t^3+4t^2-5t-2=0,t\ne 0$
Clearly $t=1$ is a solution.
So, $(t-1)(3t^2+7t+2)=0$
$\Rightarrow (t-1)(t+2)(3t+1)=0$
$\Rightarrow t=1,-2,\frac{-1}{3}$
$\therefore {\log }_{3}x=1,-2,\frac{-1}{3}$
$\Rightarrow x=3,\frac{1}{9},\frac{1}{\sqrt[3]{33}}$

Clearly, $x$ takes two rational values and one irrational value.