Question

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

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 irrational value
one positive integral value
two positive rational values

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

is valid when $x>0$

Taking logarithm with base $3$ on both sides, we get
$\left[\frac{3}{4}\right({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}]{\log }_{3}x={\log }_3\sqrt{3}={\log }_3{3}^{1/2}=\frac{1}{2},$

$[\because \log a^x=x\log a\text{\&}{\log }_{a}a=1]$

$\Rightarrow ({\log }_{3}x-1)({\log }_{3}x+2)(3{\log }_{3}x+1)=0$

$\Rightarrow {\log }_{3}x=1,$ or $-2,$ or $-\frac{1}{3}$

$\Rightarrow x=3,$ or $\frac{1}{9},$ or $\frac{1}{\sqrt[3]{33}}$
Ans: A,B.C