How do you solve \[{e^{2x + 1}} = 27\] ?

Hint: To solve the given equation, take natural logarithm on both the sides of the equation to remove the variable from the exponent, as Logarithmic functions are the inverses of exponential functions hence by this, we can get the value of \[x\].

Complete step by step solution:
Let us write the given equation
\[{e^{2x + 1}} = 27\]…………………….. 1
To solve this equation, take natural logarithm on both the sides of the equation 1 i.e.,
\[\ln \left( {{e^{2x + 1}}} \right) = \ln \left( {27} \right)\] ……………………… 2
Expand the LHS part by moving \[2x + 1\] outside the logarithm of equation 2, hence we get
\[2x + 1\ln \left( e \right) = \ln \left( {27} \right)\]
As we know the logarithm of function ‘\[e\]’ is 1, hence substituting this value in above equation
\[2x + 1\left( 1 \right) = \ln \left( {27} \right)\]
Which implies
\[2x + 1 = \ln \left( {27} \right)\] ………………………. 3
In which the value of \[\ln \left( e \right)\]= 1.
Equation 3 can be written as
\[2x + 1 = \ln {3^3}\]
\[2x + 1 = 3\ln 3\] ………………………… 4
As we need to find the value of \[x\], simplifying the terms of equation 4 we get
\[x = \dfrac{1}{2}\left( {3.\ln 3 - 1} \right)\]

Additional information:
Rules of Logarithms:
The logarithm of a positive real number can be negative, zero or positive.
Logarithmic values of a given number are different for different bases.
Logarithms to the base a 10 are referred to as common logarithms. When a logarithm is written without a subscript base, we assume the base to be 10.
Logarithms to the base ‘e’ are called natural logarithms. The constant e is approximated as 2.7183.
Natural logarithms are expressed as ln x which is the same as log e.

Note: The key point to find the given equation is that when the equation consists of exponential terms, just take natural logarithm on both the sides of the equation as to solve for the value of \[x\] we need to remove the variable from the exponent by taking ln of the function. As Logarithmic functions are the inverses of exponential functions.