How do you solve \[{{e}^{4x}}+4{{e}^{2x}}+\left( -21 \right)=0\]?

Hint: In the given question, we have been asked to find the value of ‘x’ and it is given that \[{{e}^{4x}}+4{{e}^{2x}}+\left( -21 \right)=0\]. In order to solve the question, first we need to substitute the value of\[{{e}^{2x}}=u\], then it will become a quadratic equation. Later we solve the quadratic equation and replace the ‘u’ with\[{{e}^{2x}}\]. Then simplifying the further equation, we will get our required solution.

Complete step by step solution:
We have given that,
\[\Rightarrow {{e}^{4x}}+4{{e}^{2x}}+\left( -21 \right)=0\]
Simplifying the above equation, we get
\[\Rightarrow {{e}^{4x}}+4{{e}^{2x}}-21=0\]
Substitute \[{{e}^{2x}}=u\] in the above equation, we get
As we know that \[{{e}^{4x}}={{\left( {{e}^{2x}} \right)}^{2}}\]
\[\Rightarrow {{u}^{2}}+4u-21=0\]
Splitting the middle term, we get
\[\Rightarrow {{u}^{2}}+7u-3u-21=0\]
Taking out common factor by making pairs, we get
\[\Rightarrow u\left( u+7 \right)-3\left( u+7 \right)=0\]
Taking away the common factors, we get
\[\Rightarrow \left( u+7 \right)\left( u-3 \right)=0\]
Equating each common factor equals to 0, we get
\[\Rightarrow \left( u+7 \right)=0\] And \[\left( u-3 \right)=0\]
Now, solving
\[\Rightarrow \left( u+7 \right)=0\]
Subtracting 7 from both the sides of the equation, we get
\[\Rightarrow u+7-7=0-7\]
Simplifying the above equation, we get
\[\Rightarrow u=-7\]
Now replacing the ‘u’ with \[{{e}^{2x}}\], we get
\[\Rightarrow {{e}^{2x}}=-7\]
Taking ln to both the side of the equation, we get
\[\Rightarrow \ln {{e}^{2x}}=\ln \left( -7 \right)\]
\[\Rightarrow 2x\ln e=\ln \left( -7 \right)\]
The value of ln e = 1
Putting this value, we get
\[\Rightarrow 2x=\ln \left( -7 \right)\]
Dividing both the side of the equation by 2, we get
\[\therefore x=\dfrac{\ln \left( -7 \right)}{2}=-\dfrac{\ln 7}{2}\]
Now solving,
\[\Rightarrow \left( u-3 \right)=0\]
Adding 3 to both the sides of the equation, we get
\[\Rightarrow u-3+3=0+3\]
\[\Rightarrow u=3\]
Now replacing the ‘u’ with\[{{e}^{2x}}\], we get
\[\Rightarrow {{e}^{2x}}=3\]
Taking ln to both the side of the equation, we get
\[\Rightarrow \ln {{e}^{2x}}=\ln 3\]
\[\Rightarrow 2x\ln e=\ln 3\]
The value of ln e = 1
Putting this value, we get
\[\Rightarrow 2x=\ln 3\]
Dividing both the side of the equation by 2, we get
\[\therefore x=\dfrac{\ln 3}{2}\]
Thus, the possible values of ‘x’ are \[\dfrac{\ln \left( -7 \right)}{2}\] and\[\dfrac{\ln 3}{2}\]. But logarithmic function cannot take negative arguments as it will not generate a real solution.
Therefore, The only real solution is \[x=\dfrac{\ln 3}{2}\].

Note: In the given question, we need to find the value of ‘x’. To solve these types of questions, we used the basic formulas of logarithm. Students should always require to keep in mind all the formulae of logarithmic function for solving the question easily. After applying log formulae to the equation, we need to solve the equation in the way we solve general equations.