If an algebraic expression is given as ${{2}^{x}}{{.3}^{x+4}}={{7}^{x}}$ , then x is equal to
(a) \[\dfrac{4{{\log }_{e}}3}{{{\log }_{e}}7-{{\log }_{e}}6}\]
(b) \[\dfrac{-4{{\log }_{e}}3}{{{\log }_{e}}7-{{\log }_{e}}6}\]
(c) \[\dfrac{2{{\log }_{e}}3}{{{\log }_{e}}7-{{\log }_{e}}6}\]
(d) \[\dfrac{-2{{\log }_{e}}3}{{{\log }_{e}}7-{{\log }_{e}}6}\]

Hint: It is a complex exponential equation with different bases like 2, 3, and 7. So, here we will make a common base, and then we will take a logarithm on both sides.

Complete step-by-step solution -
In the question given, we have different values of bases and different powers on them, so we cannot just equate base on both sides. So here, first we will multiply or divide the complete equation with such a number such that one side of the equation is converted into a constant term. Thus, first we will have to simplify the exponential equation given. For this,
${{2}^{x}}{{.3}^{x+4}}={{7}^{x}}...................\left( i \right)$
Here, we are going to use an identity which is as shown:
${{a}^{x+y}}={{a}^{x}}.{{a}^{y}}$
Applying the above identity in equation (i), we get
$\Rightarrow {{2}^{x}}{{.3}^{x}}{{.3}^{4}}={{7}^{x}}...............\left( ii \right)$
Here, we are using another identity which is shown below:
${{a}^{x}}\times {{b}^{x}}={{\left( a\times b \right)}^{x}}$
Using this identity in equation (ii), we will get following
$\begin{align}
  & \Rightarrow {{\left( 2\times 3 \right)}^{x}}{{.3}^{4}}={{7}^{x}} \\
 & \Rightarrow {{6}^{x}}{{.3}^{4}}={{7}^{x}}................\left( iii \right) \\
\end{align}$
Now, we will divide both the sides of the equation with ${{6}^{x}}$ . After dividing, we will get the following:
$\begin{align}
  & \Rightarrow \dfrac{{{{6}^{x}}.{{3}^{4}}}}{{{6}^{x}}}=\dfrac{{{7}^{x}}}{{{6}^{x}}} \\
 & \Rightarrow {{3}^{4}}=\dfrac{{{7}^{x}}}{{{6}^{x}}} \\
 & \Rightarrow \dfrac{{{7}^{x}}}{{{6}^{x}}}={{3}^{4}}..............\left( iv \right) \\
\end{align}$
Here, we are using another identity as shown below:
$\dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}}$
Using the above identity in equation (iv) we get
${{\left( \dfrac{7}{6} \right)}^{x}}={{3}^{4}}.............\left( v \right)$
Now, here one side of the equation is constant and other side is power of a term. This can be solved by taking logarithm on both sides. The base of the logarithm we are going to takes is $\dfrac{7}{6}$ . Thus, after taking logarithm on both sides, we will get
${{\log }_{\left( \dfrac{7}{6} \right)}}{{\left( \dfrac{7}{6} \right)}^{x}}={{\log }_{\left( \dfrac{7}{6} \right)}}{{3}^{4}}..................\left( vi \right)$
Here the identity which we are going to use is shown below
${{\log }_{a}}{{b}^{\alpha }}=\alpha {{\log }_{a}}b$
Using the identity in equation (vi), we will get following:
$x{{\log }_{\left( \dfrac{7}{6} \right)}}\left( \dfrac{7}{6} \right)=4{{\log }_{\left( \dfrac{7}{6} \right)}}3.....................\left( vii \right)$
In this step, the identity we are going to use is:
${{\log }_{a}}a=1$
Thus, after using this identity, we will get:
$\begin{align}
  & \Rightarrow x\times 1=4{{\log }_{\left( \dfrac{7}{6} \right)}}3 \\
 & \Rightarrow x=4{{\log }_{\left( \dfrac{7}{6} \right)}}3..................\left( viii \right) \\
\end{align}$
We have got our answer but we will have to convert the above equation in terms of option given. To do this we will take the base of logarithm as ‘e’. Thus to take this base as ‘e’, we have base changing property:
${{\log }_{a}}b=\dfrac{{{\log }_{e}}b}{{{\log }_{e}}a}$
Using this identity in equation (viii) we get
$x=\dfrac{4{{\log }_{e}}3}{{{\log }_{e}}\left( \dfrac{7}{6} \right)}................\left( ix \right)$
Now, we will use another identity as shown:
${{\log }_{e}}\left( \dfrac{b}{a} \right)={{\log }_{e}}b-{{\log }_{c}}a$
After applying identity in equation (ix), we get
$x=\dfrac{4{{\log }_{e}}3}{{{\log }_{3}}7-{{\log }_{e}}6}$
This is our required answer.
Hence, option (a) is correct.

Note: We have not made the use of domain while using exponential identities because the base is positive every time. So, the necessary condition for ${{a}^{x}}$ to exist is that $a>0$ .