How do you solve \[lo{{g}_{6}}x+lo{{g}_{6}}\left( x-9 \right)=2\] ?

Hint: At first, we apply the property of logarithms that $\log a+\log b=\log ab$ . Here, we put the values of $a=x$ and $b=\left( x-9 \right)$ . Then, we reverse the logarithms to the indices. We now get a quadratic equation, which can be solved using the middle term method.

Complete step-by-step answer:
Logarithms and indices go hand in hand. Indices are a way to express what will be the output of the input “a base number raised to some power”. For example, the index “ ${{10}^{2}}$ “ just gives the output $100$ . Here, we are interested in what raising a number by a power will give. In logarithms, the thing is opposite. Logarithms are a way to express “what should be the power to which a number must be raised to make it equal to another number”. For example, the number “ $100$ “ just gives the output $2$ when the base is $10$ .
There is a property of logarithms which states that
$\log a+\log b=\log ab$ until and unless the base is same for both the logarithms.
Using a and b equal to $a=x$ and $b=\left( x-9 \right)$ in the above equation, we get,
\[\begin{align}
  & \Rightarrow lo{{g}_{6}}x+lo{{g}_{6}}\left( x-9 \right)=2 \\
 & \Rightarrow {{\log }_{6}}x\left( x-9 \right)=2 \\
\end{align}\]
We can rearrange the above equation based on the concept of logarithms. This is done by,
\[\begin{align}
  & \Rightarrow x\left( x-9 \right)={{6}^{2}} \\
 & \Rightarrow {{x}^{2}}-9x-36=0 \\
 & \Rightarrow {{x}^{2}}-12x+3x-36=0 \\
 & \Rightarrow x\left( x-12 \right)+3\left( x-12 \right)=0 \\
 & \Rightarrow \left( x-12 \right)\left( x+3 \right)=0 \\
 & \Rightarrow x=12,-3 \\
\end{align}\]
Therefore, we can conclude that the solution for the given equation is \[x=12,-3\] .

Note: These problems require the knowledge of both logarithms and quadratic equations. So, as this requires both knowledge, we must be extra cautious over here. For the logarithms, we must check if the bases of the logarithms are the same or not before applying the property. Also, the quadratic can be solved by using the Sridhar Acharya formula.