How do you solve the logarithmic equation ${{\log }_{3}}x-2{{\log }_{x}}3=1$ ?

Hint: Problems of this type can be solved by rewriting the logarithmic terms by using the change of base formula of logarithms. We elaborate the terms as a quotient of logarithmic terms having a common positive base. Thus, the equation becomes a quadratic equation which is further solved by factorization method.

Complete step-by-step solution:
The equation we have is
${{\log }_{3}}x-2{{\log }_{x}}3=1$
As both the logarithmic terms have different bases, we use the change of base formula of logarithm.
According to the Change of base formula of logarithm
${{\log }_{b}}a=\dfrac{\log a}{\log b}$
Therefore, we rewrite the term ${{\log }_{3}}x$ as $\dfrac{\log x}{\log 3}$ and the term ${{\log }_{x}}3$ as $\dfrac{\log 3}{\log x}$
Substituting the above terms in the given equation we get
$\Rightarrow \dfrac{\log x}{\log 3}-2\dfrac{\log 3}{\log x}=1$
We assume that $\dfrac{\log x}{\log 3}=y$
Substituting the above expression with $y$ in the above equation we get
\[\Rightarrow y-\dfrac{2}{y}=1\]
Multiplying the above equation with $y$ to both the sides we get
$\Rightarrow {{y}^{2}}-2=y$
Further simplifying the above equation, we get
$\Rightarrow {{y}^{2}}-y-2=0$
The above equation has become a quadratic equation. So, we can rewrite the coefficient of $y$ as
$\left( 2-1 \right)=1$ , as $2$ and $1$ are the multipliers of $2$
Hence, the equation can be written as
$\Rightarrow {{y}^{2}}-\left( 2-1 \right)y-2=0$
We now apply the distribution on the middle term of the expression and get
$\Rightarrow {{y}^{2}}-2y+y-2=0$
We can now take $y$ common from ${{y}^{2}}-2y$ and $1$ common from the rest terms as
$\Rightarrow y\left( y-2 \right)+1\left( y-2 \right)=0$
Again, we can see that the term $\left( y-2 \right)$ can be taken as common from the entire expression as
$\Rightarrow \left( y-2 \right)\left( y+1 \right)=0$
Hence,
$y-2=0$
$\Rightarrow y=2$
Substituting the expression of $y$ we get
$\Rightarrow \dfrac{\log x}{\log 3}=2$
$\Rightarrow \log x=2\log 3$
$\Rightarrow \log x=\log 9$
$\Rightarrow x=9$
Again, $y+1=0$
$\Rightarrow y=-1$
Substituting the expression of $y$ we get
$\Rightarrow \dfrac{\log x}{\log 3}=-1$
$\Rightarrow \log x=-\log 3$
$\Rightarrow \log x=\log \dfrac{1}{3}$
$\Rightarrow x=\dfrac{1}{3}$
Therefore, we conclude that the solutions of the given equation are $x=9,\dfrac{1}{3}$.

Note: While changing the base of the logarithmic terms we must be very careful so that mistakes are avoided. Also, the quadratic equation we got can be solved by other methods, such as forming a square method or by directly using Sridhar Acharya's formula.