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The equation to solve for \( x \) is:

\[
e^{2x} + 3e^x - 10 = 0
\]

To solve this equation, let's set \( u = e^x \). This substitution transforms the equation into a quadratic form:

\[
u^2 + 3u - 10 = 0
\]

Now, solve this quadratic equation using the quadratic formula:

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

where \( a = 1 \), \( b = 3 \), and \( c = -10 \):

\[
u = \frac{-3 \pm \sqrt{3^2 - 4(1)(-10)}}{2(1)} = \frac{-3 \pm \sqrt{9 + 40}}{2} = \frac{-3 \pm \sqrt{49}}{2} = \frac{-3 \pm 7}{2}
\]

Thus, we get two solutions for \( u \):

\[
u = \frac{-3 + 7}{2} = 2 \quad \text{or} \quad u = \frac{-3 - 7}{2} = -5
\]

Now, recall that \( u = e^x \), so we solve for \( x \) in both cases:

1. \( e^x = 2 \) gives \( x = \ln(2) \)
2. \( e^x = -5 \) is not possible because the exponential function \( e^x \) is always positive.

Therefore, the only valid solution is \( x = \ln(2) \).

Thus, the correct answer is **\(\ln(2)\)**.

Would you like any further details or clarification? Here are 5 related questions you might explore:
1. How can the quadratic formula be derived?
2. What is the domain of the exponential function?
3. Why is there no solution for \( e^x = -5 \)?
4. How do logarithms relate to exponential equations?
5. Can all quadratic equations be solved using substitution?

**Tip:** When solving exponential equations, always check if the solution is within the domain of the exponential function!

Learn how to solve the exponential equation e^(2x) + 3e^x - 10 = 0 using substitution and the quadratic formula. This solution involves key concepts such as exponential functions and logarithms. Suitable for students in Grades 10-12.

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