To solve the exponential equation [tex]\(e^{2x + 1} = 7\)[/tex], we will follow these steps:
1. Take the natural logarithm (ln) on both sides of the equation:
[tex]\[
\ln(e^{2x + 1}) = \ln(7)
\][/tex]
2. Apply the property of logarithms that states [tex]\(\ln(e^y) = y \cdot \ln(e)\)[/tex]. Because [tex]\(\ln(e) = 1\)[/tex], this simplifies to [tex]\( y \)[/tex]:
[tex]\[
2x + 1 = \ln(7)
\][/tex]
3. Isolate the term involving [tex]\(x\)[/tex]. To do this, subtract 1 from both sides of the equation:
[tex]\[
2x = \ln(7) - 1
\][/tex]
4. Solve for [tex]\(x\)[/tex] by dividing both sides by 2:
[tex]\[
x = \frac{\ln(7) - 1}{2}
\][/tex]
To express this value in numerical form correct to four decimal places, we compute it as follows:
[tex]\[
x \approx 0.4730
\][/tex]
Thus, the solution to the equation [tex]\(e^{2x + 1} = 7\)[/tex] is:
[tex]\[
x \approx 0.4730
\][/tex]
