How do you solve ${{e}^{3x}}=12$ ?

Hint: For answering this question we will consider the given equation and apply logarithm to base $e$ on both sides. And use our basic logarithm concept and apply the formulae ${{\log }_{a}}\left( {{a}^{n}} \right)=n$ and solve the given expression and conclude the answer.

Complete step by step answer:
Now considering from the question we have the expression as given ${{e}^{3x}}=12$ .
The basic logarithm concept gives us ${{\log }_{a}}x=y\Rightarrow {{a}^{y}}=x$.
By applying logarithm of base $e$ on both sides we will have ${{\log }_{e}}\left( {{e}^{3x}} \right)={{\log }_{e}}12$ .
By using this basic concept we can derive the formulae ${{\log }_{a}}\left( {{a}^{n}} \right)=n$ as follows.
Initially by comparing we can say $x={{a}^{n}}$ by using this we will have ${{a}^{n}}={{a}^{y}}$ then we can conclude that $y=n$ .
As from the basic concepts of logarithm we are aware that ${{\log }_{a}}\left( {{a}^{n}} \right)=n$ so we will have $3x={{\log }_{e}}\left( 12 \right)$ .
By further simplifying this using basic arithmetic calculations we will have $x=\dfrac{{{\log }_{e}}\left( 12 \right)}{3}$ .

Hence we can conclude that the value of $x$ is $\dfrac{{{\log }_{e}}\left( 12 \right)}{3}$

Note: We should be sure with our calculations and logarithm concepts for answering this question. The basic logarithmic concept gives us some basic formulae which are ${{\log }_{a}}\left( {{a}^{n}} \right)=n$ , $\log a-\log b=\log \left( \dfrac{a}{b} \right)$ , $\log a+\log b=\log \left( ab \right)$ , ${{a}^{{{\log }_{a}}\left( {{a}^{n}} \right)}}={{a}^{n}}$ , ${{\log }_{a}}x=y\Rightarrow {{a}^{y}}=x$ . As shown above we can derive all these formulae and we can also derive many other formulae which we have not discussed here. These formulae can be applied in questions of this type and efficiently solve them within less time so it is suggested to keep them in mind.