Question

The equation $\left(\sqrt{1+{\log }_x\sqrt{27}}\right){\log }_{3}x+1=0$ has

• A. no integral solution
• B. one irrational solution
• C. two real solutions
• D. no prime solution

Answer 1

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

The correct options are
A no integral solution
B one irrational solution
C two real solutions
D no prime solution

Clearly, $x>0,x\ne 1$
$\left(\sqrt{1+{\log }_x\sqrt{27}}\right){\log }_{3}x+1=0$
$\Rightarrow \left(\sqrt{1+\frac{3}{2}{\log }_{x}3}\right){\log }_{3}x+1=0\Rightarrow \sqrt{1+\frac{3}{2{\log }_{3}x}}=-\frac{1}{{\log }_{3}x}$

Let ${\log }_{3}x=t$
Then $\sqrt{1+\frac{3}{2t}}=-\frac{1}{t},t\ne 0$
$\Rightarrow 1+\frac{3}{2t}=\frac{1}{t^2}\Rightarrow 2t^2+3t-2=0\Rightarrow t=-2,\frac{1}{2}\Rightarrow {\log }_{3}x=-2,\frac{1}{2}\therefore x=\frac{1}{9},\sqrt{3}$