To solve the equation [tex]\( e^{2x + 1} = 33 \)[/tex] for [tex]\( x \)[/tex], we can follow these steps:
1. Isolate the exponential term:
Given the equation [tex]\( e^{2x + 1} = 33 \)[/tex], we want to isolate the exponential expression. We'll begin by taking the natural logarithm on both sides of the equation to eliminate the exponential function. The natural logarithm ([tex]\( \ln \)[/tex]) and the exponential function ([tex]\( e \)[/tex]) are inverse functions.
2. Apply the natural logarithm:
[tex]\[
\ln(e^{2x + 1}) = \ln(33)
\][/tex]
3. Simplify the left side:
Using the property of logarithms that [tex]\( \ln(e^y) = y \)[/tex], we simplify the left-hand side:
[tex]\[
2x + 1 = \ln(33)
\][/tex]
4. Isolate [tex]\( x \)[/tex]:
Rearrange the equation to solve for [tex]\( x \)[/tex]. First, subtract 1 from both sides:
[tex]\[
2x = \ln(33) - 1
\][/tex]
Then, divide both sides by 2:
[tex]\[
x = \frac{\ln(33) - 1}{2}
\][/tex]
Simplifying this expression:
[tex]\[
x = \frac{\ln(33) - 1}{2}
\][/tex]
So, the solution to the equation [tex]\( e^{2x + 1} = 33 \)[/tex] for [tex]\( x \)[/tex] is:
[tex]\[
x = -\frac{1}{2} + \frac{\ln(33)}{2}
\][/tex]
