Question

If $x^{\frac{3}{4}(lo{g}_{3}x{)}^2+lo{g}_{3}x-\frac{5}{4}}=\sqrt{3}$ 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)

The correct options are
A One positive integral value
B One irrational value
C Two positive rational values

$x^{\frac{3}{4}(lo{g}_{3}x{)}^2+lo{g}_{3}x-\frac{5}{4}}=\sqrt{3}=3^{\frac{1}{2}}.$
There is a possibility of a solution x = 3

For this value, $LHS=3^{\frac{3}{4}.1^2+1-\left(\frac{5}{4}\right)}=3^{\frac{2}{4}}=3^{\frac{1}{2}}=RHS.$
$\therefore x=3$ is a solution, which is a integer.
Next, $\left[\frac{3}{4}\right(lo{g}_{3}x{)}^2+lo{g}_{3}x-\frac{5}{4}]lo{g}_{3}x=\frac{1}{2}$
$\Rightarrow \left[3\right(lo{g}_{3}x{)}^2+4lo{g}_{3}x-5]lo{g}_{3}x-2=0\Rightarrow 3t^3+4t^2-5t-2=0,[t=lo{g}_{3}x]\Rightarrow 3t^3-3t^2+7t^2-7t+2t-2=0\Rightarrow (3t^2+7t+2\left)\right(t-1)=0\Rightarrow (3t+1\left)\right(t+2\left)\right(t-1)=0$
$\Rightarrow t=1,-2,-\frac{1}{3}\Rightarrow lo{g}_{3}x=1,-2,-\frac{1}{3}$
$\Rightarrow x=3^1,3^{-2},3^{\frac{-1}{3}};\therefore x=3,\frac{1}{9},\frac{1}{\sqrt[3]{33}}$
Thus, there is one +ve integral value, one irrational value, two positive rational values.