How do you solve \[2\log \left( 3 \right)+\log \left( x \right)=\log \left( 36 \right)\] ?

Hint: The above equation can be done by simply recalling the basic form and formulae of logarithms. The basic form of logarithm is expressed as:
\[{{\log }_{b}}{{a}^{x}}=x{{\log }_{b}}a\] .Here \[b\] is the base, \[a\] is the exponent and \[x\] is the power.
Now, if \[a=b\] , \[{{\log }_{a}}{{a}^{x}}=x{{\log }_{a}}a=x\] as we know \[{{\log }_{a}}a=1\]

Complete step by step answer:
The basic formulae for logarithms are:
\[\begin{align}
  & \log a+\log b=\log \left( ab \right) \\
 & \log a-\log b=\log \left( \dfrac{a}{b} \right) \\
\end{align}\]
When no base has been mentioned in the problem, we assume the base to be as \[10\] . We also need to remember that, if \[{{\log }_{10}}a=c\] then we can easily find out the value of \[a\] as, \[a={{10}^{c}}\] .
Applying the second equation in the question, which states,
\[\log a-\log b=\log \left( \dfrac{a}{b} \right)\]
Thus, we can write,
\[2\log \left( 3 \right)+\log \left( x \right)=\log \left( 36 \right)\]
Using \[{{\log }_{b}}{{a}^{x}}=x{{\log }_{b}}a\] law, we can write the above equation as,
\[\log \left( {{3}^{2}} \right)+\log \left( x \right)=\log \left( 36 \right)\]
Evaluating we can write it as,
\[\log \left( 9 \right)+\log \left( x \right)=\log \left( 36 \right)\]
Now, using the logarithmic rules in the above equation on both the L.H.S and R.H.S we get,
\[\log \left( 9x \right)=\log \left( 36 \right)\]
Now, taking antilogarithm on both sides, we get the following equation,
\[9x=36\]
Now, to find the value of \[x\] we need to divide the right hand side of the equation with the coefficient of \[x\] . Here the coefficient of \[x\] is \[9\] .
So we get,
\[x=\dfrac{36}{9}\]
Dividing, we get,
\[x=4\]
This is a valid solution, as if we put \[x=4\] in our original given equation, it satisfies both the L.H.S and R.H.S.
Following the above steps, we can very easily find our required solution. Our final answer is hence \[x=4\] .

Note:
We need to keep in mind of all the logarithmic formulae and relations for smooth solving of these types of questions. We also have to keep in mind that \[\log 0\] is undefined and hence to avoid such kinds of complications in our solution.