Question

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

• A. One positive integral value
• B. One irrational value
• C. two positive rational value
• D. All of these

Answer 1

The correct option is D

All of these

$x^{\left(\frac{3}{4}\right({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4})}=3^{\frac{1}{2}}$ -----------(1)

Since logarithum number is given on the base 3

There is possibility of solution of power of 3

Let's check for x = 3

LHS = $3^{\frac{3}{4}}({\log }_{3}3{)}^2+{\log }_{3}3-\frac{5}{4}$

= $3^{\frac{3}{4}}\left(1\right)+1-\frac{5}{4}$ = $3^{\frac{3}{4}}+1-\frac{5}{4}$ = $3^{\frac{1}{2}}$ = $\sqrt{3}$ = RHS

So,x=3 is a solution of the equation which is a positive integer.

But we need to find all the roots of the above given equation

Now, solve it from standard method

Taking log on both side of equation 1 at base 3

${\log }_3\left\{x^{\frac{3}{4}({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}}\right\}={\log }_3{3}^{\frac{1}{2}}$

$\left(\frac{3}{4}\right({\log }_{3}x{)}^2+{\log }_{3}x-\frac{5}{4}){\log }_{3}x=\frac{1}{2}$

let $y=lo{g}_{3}x$
$\Rightarrow (\frac{3}{4}{y}^2+y-\frac{5}{4})y=\frac{1}{2}$
$\Rightarrow 3y^3+4y^2-5y-2=0$
Clearly y =1 satisties the equation

$\therefore 3y^2+7y+2=0$
$\Rightarrow (3y+1)(y+2)=0$
$\Rightarrow y=\frac{-1}{3},-2$
When $y=\frac{-1}{3}lo{g}_{3}x=\frac{-1}{3}$
$\Rightarrow x=3^{-1/3}=\frac{1}{3\sqrt{3}}$ is irrational value
when $y=-2lo{g}_{3}x=-2$
$\Rightarrow x=3^{-2}=\frac{1}{9}$ rational value