How do you solve \[2{\log _3}\left( {x + 4} \right) = {\log _3}\left( 9 \right) + 2\] ?

Hint: In the given question, we are required to solve a logarithmic equation provided to us in the problem itself. We can also make use of properties of logarithms to make the calculation part easier and less time-consuming. Such questions require thorough knowledge about the basics of applications of logarithms.

Complete step by step answer:
We have, \[2{\log _3}\left( {x + 4} \right) = {\log _3}\left( 9 \right) + 2\]
Now, we know that \[{\log _3}\left( 9 \right) = 2\] as \[{\log _3}\left( 9 \right) = {\log _3}\left( {{3^2}} \right)\], so substituting the value of \[{\log _3}\left( 9 \right)\], we get,
\[ \Rightarrow 2{\log _3}\left( {x + 4} \right) = 2 + 2\]
Simplifying the right side of the equation further, we get,
\[ \Rightarrow 2{\log _3}\left( {x + 4} \right) = 4\]
Dividing both sides of the equation by \[2\], we get,
\[ \Rightarrow {\log _3}\left( {x + 4} \right) = 2\]
Now, the exponential form ${a^n} = b$where base $ = a$, exponent$ = n$, value$ = b$ is written as ${\log _a}b = n$ in logarithmic form, read as log of ‘b’ to base ‘a’ is equal to ‘n’.
Therefore, the logarithmic form, ${\log _a}b = n$, read as log of ‘b’ to base ‘a’ is equal to ‘n’ can be written as ${a^n} = b$ in exponential form.
So, we get,
\[ \Rightarrow \left( {x + 4} \right) = {3^2}\]
We know that ${3^2} = 9$, we get,
\[ \Rightarrow \left( {x + 4} \right) = 9\]
Subtracting $4$ from both sides of the equation, we get,
\[ \Rightarrow x = 5\]

So, we get the value of x is \[5\].

Note: Logarithmic properties are of much use in such logarithmic equations. Such problems require thorough knowledge of properties of logarithms and their applications. Besides the concepts and applications of logarithms, we need to have a strong grip of algebraic rules and identities in order to correctly solve such types of questions.