How do you solve \[{{3}^{x+4}}={{2}^{1-3x}}\]?

Hint: This question belongs to the topic of pre-calculus. In this question, for solving we will take log to the both sides of the equation. After that, we will take out the powers from both sides of the equation. After that, we will expand the terms by multiplying the term inside the bracket with the term outside the bracket. After that, we will solve the further equation to get the value of x.

Complete step by step answer:
Let us solve this question.
In this question, we have to solve the term \[{{3}^{x+4}}={{2}^{1-3x}}\]. Or, we can say that we have to solve the equation \[{{3}^{x+4}}={{2}^{1-3x}}\] and find the value of x.
The equation we have to solve is
\[{{3}^{x+4}}={{2}^{1-3x}}\]
After taking log base e (that is \[{{\log }_{e}}\] or \[\ln \]) to the both side of the equation, we get
\[\Rightarrow \ln \left( {{3}^{x+4}} \right)=\ln \left( {{2}^{1-3x}} \right)\]
As we know that \[\ln {{x}^{a}}\] can also be written as \[a\ln x\]. So, using this we can write the above equation as
\[\Rightarrow \left( x+4 \right)\ln 3=\left( 1-3x \right)\ln 2\]
The above equation can also be written as
\[\Rightarrow x\ln 3+4\ln 3=\ln 2-3x\ln 2\]
Taking all the terms having coefficients as x to the left side of equation and the rest of the terms to the right side of the equation, we get
\[\Rightarrow x\ln 3+3x\ln 2=\ln 2-4\ln 3\]
The above can also be written as
\[\Rightarrow x\left( \ln 3+3\ln 2 \right)=\ln 2-4\ln 3\]
Using the values as
\[\ln 2=0.693\]
\[\ln 3=1.098\] in the above equation, we get
\[\Rightarrow x\left( 1.098+3\times 0.693 \right)=0.693-4\times 1.098\]
The above can also be written as
\[\Rightarrow x\left( 1.098+2.079 \right)=0.693-4.392\]
The above equation can also be written as
\[\Rightarrow x\left( 3.177 \right)=-3.699\]
The above can also be written as
\[\Rightarrow x=-\dfrac{3.699}{3.177}\]
The above can also be written as
\[\Rightarrow x=-1.164\]
Hence, we have solved the equation \[{{3}^{x+4}}={{2}^{1-3x}}\] and have got the value of x as -1.164

Note:
As we can see that this question is from the topic of pre-calculus, so that we should have a better knowledge in that topic to solve this type of question easily.
Always remember that ‘\[\log \] base e’ can also be written as\[\ln \].
Remember that \[\ln {{x}^{a}}\] can also be written as \[a\ln x\], or we can say that \[\ln {{x}^{a}}=a\ln x\]
Remember the following values:
\[\ln 2=0.693\]
\[\ln 2=0.693\]
 Always remember the above values and formulas to solve this type of question easily.