To solve the equation \( e^{2x} - 11e^x + 28 = 0 \), let's start by making a substitution. Let \( u = e^x \). This simplifies the equation to:

\[
u^2 - 11u + 28 = 0
\]

This is now a quadratic equation in terms of \( u \). We can solve this using the quadratic formula:

\[
u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

Here, \( a = 1 \), \( b = -11 \), and \( c = 28 \). Plugging these values into the quadratic formula:

\[
u = \frac{-(-11) \pm \sqrt{(-11)^2 - 4(1)(28)}}{2(1)}
\]

\[
u = \frac{11 \pm \sqrt{121 - 112}}{2}
\]

\[
u = \frac{11 \pm \sqrt{9}}{2}
\]

\[
u = \frac{11 \pm 3}{2}
\]

This gives us two possible values for \( u \):

\[
u_1 = \frac{11 + 3}{2} = 7
\]
\[
u_2 = \frac{11 - 3}{2} = 4
\]

Now, recall that \( u = e^x \), so we have:

\[
e^x = 7 \quad \text{or} \quad e^x = 4
\]

Taking the natural logarithm of both sides to solve for \( x \):

\[
x = \ln(7) \quad \text{or} \quad x = \ln(4)
\]

Thus, the solutions to the equation are:

\[
x = \ln(7) \quad \text{or} \quad x = \ln(4)
\]

Would you like more details on this solution or have any questions? 

Here are 5 related questions to explore further:

1. How would you solve a similar equation if the coefficients were different?
2. What are the properties of exponential functions that are useful in solving such equations?
3. How can you verify that \( x = \ln(7) \) and \( x = \ln(4) \) are correct solutions?
4. How would the solution change if the equation had a negative constant instead of \( +28 \)?
5. Can you solve this equation using graphing methods? How would you do it?

**Tip:** For exponential equations, substitution is a powerful technique to simplify the problem. Always look for patterns that can reduce complex expressions to quadratic forms.

Learn how to solve the exponential equation e^(2x) - 11e^x + 28 = 0 step-by-step with detailed explanations. This problem involves substituting variables, applying the quadratic formula, and understanding exponential functions and natural logarithms.

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