To solve the equation \( e^{2x} + 3e^x - 28 = 0 \), we can use a substitution. Let \( y = e^x \). Then \( e^{2x} = y^2 \), and our equation becomes:
\[ y^2 + 3y - 28 = 0 \]
This is a quadratic equation in standard form \( ay^2 + by + c = 0 \), where \( a = 1, b = 3, c = -28 \).
We can use the quadratic formula:
\[ y = \frac{{-b \pm \sqrt{{b^2 - 4ac}}}}{2a} \]
Plugging in the values:
\[ y = \frac{{-3 \pm \sqrt{{3^2 - 4(1)(-28)}}}}{2(1)} \] \[ y = \frac{{-3 \pm \sqrt{{9 + 112}}}}{2} \] \[ y = \frac{{-3 \pm \sqrt{121}}}{2} \] \[ y = \frac{{-3 \pm 11}}{2} \]
Calculating the two possible values for \( y \):
- \( y = \frac{{-3 + 11}}{2} = \frac{8}{2} = 4 \)
- \( y = \frac{{-3 - 11}}{2} = \frac{-14}{2} = -7 \)
Since \( y = e^x \) and \( e^x \) can only be positive, we discard \( y = -7 \) and keep \( y = 4 \).
Now, substituting back for \( y \):
\[ e^x = 4 \]
Taking the natural logarithm of both sides:
\[ x = \ln(4) \]
Thus, the solution to the original equation is:
\[ x = \ln(4) \]
The other proposed solutions (ln(-7) or ln(-4) or ln(7)) do not apply, as logarithms of negative numbers are undefined.
Final Answer: \( x = \ln(4) \)