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For x, a>0 the roots of the equation ${{\log }_{ax}}a+{{\log }_{x}}{{a}^{2}}+{{\log }_{{{a}^{2}}x}}{{a}^{3}}=0$ is (are) given by
(a) ${{a}^{-\dfrac{4}{3}}}$
(b) ${{a}^{-\dfrac{3}{4}}}$
(c) ${{a}^{-\dfrac{1}{2}}}$
(d) ${{a}^{-\dfrac{2}{3}}}$

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Hint: First convert out given expression using the formula, ${{\log }_{a}}x=\dfrac{\log a}{\log x} ,\log (cd)=\log c+\log d,\log \left( {{d}^{n}} \right)=n\log d$. Then divide the whole expression by common term. Then transform the obtained expression using ${{\log }_{a}}x=t$, then find the value of ‘t’ and then transform it to get values of x in terms of a.

Complete step-by-step answer:
We are given that for x, a > 0 we have to find the roots of equation
${{\log }_{ax}}a+{{\log }_{x}}{{a}^{2}}+{{\log }_{{{a}^{2}}x}}{{a}^{3}}=0.............\left( i \right)$
Now to proceed the equation we have to use formula such as,
${{\log }_{c}}d=\dfrac{\log d}{\log c}$
By using this we can transform equation (i) as,
$\dfrac{\log a}{\log ax}+\dfrac{\log {{a}^{2}}}{\log x}+\dfrac{\log {{a}^{3}}}{\log {{a}^{2}}x}=0............\left( ii \right)$
Now we will use the formula such as,
log(cd) = log c +log d
By using this we can write equation (ii) as,
$\dfrac{\log a}{\log a+\log x}+\dfrac{\log {{a}^{2}}}{\log x}+\dfrac{\log {{a}^{3}}}{\log {{a}^{2}}+\log x}=0............\left( iii \right)$
Now we will use the formula such as,
$\log \left( {{d}^{n}} \right)=n\log d$
By using this we can write the equation (iii) as,
$\dfrac{\log a}{\log a+\log x}+\dfrac{2\log a}{\log x}+\dfrac{3\log a}{2\log a+\log x}=0............\left( iv \right)$
Now we will divide log a from both numerator and denominator of the term of equation (iv) we get,
$\dfrac{1}{1+\dfrac{\log x}{\log a}}+\dfrac{2}{\dfrac{\log x}{\log a}}+\dfrac{3}{2+\dfrac{\log x}{\log a}}=0...............\left( v \right)$
In equation (v) we will use the formula, $\dfrac{\log c}{\log d}={{\log }_{d}}c$ we get,
$\dfrac{1}{1+{{\log }_{a}}x}+\dfrac{2}{{{\log }_{a}}x}+\dfrac{3}{2+{{\log }_{a}}x}=0..............\left( vi \right)$
Now we will substitute ${{\log }_{a}}x=t$ , we will transform equation (vi) as
\[\dfrac{1}{1+t}+\dfrac{2}{t}+\dfrac{3}{2+t}=0............\left( vii \right)\]
Taking LCM in (vii) we get,
$\dfrac{t\left( 2+t \right)+2\left( 1+t \right)(2+t)+3t\left( t+1 \right)}{\left( 1+t \right)t\left( 2+t \right)}=0$
On cross multiplication we get,
t(2 + t) + 2(1 + t)(2+t) +3t (t + 1) = 0
On further simplification we get,
$\begin{align}
  & 2t+{{t}^{2}}+2(2+t+2t+{{t}^{2}})+3{{t}^{2}}+3t=0 \\
 & \Rightarrow 2t+{{t}^{2}}+4+2t+4t+2{{t}^{2}}+3{{t}^{2}}+3t=0 \\
\end{align}$
On simplification we get,
$6{{t}^{2}}+11t+4=0$
This is a quadratic equation. We will solve it by splitting the middle term, we get
$\begin{align}
 & 6{{t}^{2}}+8t+3t+4=0 \\
 & \Rightarrow 2t\left( 3t+4 \right)+1\left( 3t+4 \right)=0 \\
 & \Rightarrow (3t+4)(2t+1)=0 \\
 & \Rightarrow 3t+4=0,2t+1=0 \\
 & \Rightarrow t=-\dfrac{4}{3},t=-\dfrac{1}{2} \\
\end{align}$
So, finally the value for $t=\dfrac{-4}{3},\dfrac{-1}{2}$.
We had assumed,
${{\log }_{a}}x=t$
Substituting the value of ‘t’, we get
${{\log }_{a}}x=\dfrac{-4}{3}$ and ${{\log }_{a}}x=-\dfrac{1}{2}$
Now we will use the transformation that is, ${{\log }_{b}}a=c\Rightarrow a={{b}^{c}}$. By using this in above equation, we get
$x={{a}^{-\dfrac{4}{3}}}$ and $x={{a}^{\dfrac{-1}{2}}}$ respectively.
Therefore, the correct answer is option (a) and (c).

Note: Students should be careful while calculating and finding values of ${{\log }_{a}}x$ also using transformation of changing ${{\log }_{a}}x=t$as $x={{a}^{t}}$.
Another way to solve the quadratic equation is using the formula, $t=\dfrac{-b\pm \sqrt{{{b}^{2}}-4ac}}{2a}$.
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