To solve the equation $e^{2x} - 6e^x + 8 = 0$, we can use substitution. Let's proceed step by step:

Step 1: Substitute

Let $u = e^x$. This means $u^2 = e^{2x}$. Now, substitute this into the equation: $u^2 - 6u + 8 = 0$ This is now a quadratic equation in terms of $u$.

Step 2: Solve the quadratic equation

To solve the quadratic equation $u^2 - 6u + 8 = 0$, we can factor it: $u^2 - 6u + 8 = (u - 4)(u - 2) = 0$

Thus, the solutions for $u$ are: $u = 4 \quad \text{or} \quad u = 2$

Step 3: Solve for $x$

Recall that $u = e^x$. So now we solve for $x$:

1. If $e^x = 4$, take the natural logarithm of both sides: $x = \ln(4)$

2. If $e^x = 2$, take the natural logarithm of both sides: $x = \ln(2)$

Final Answer:

The solutions are: $x = \ln(4) \quad \text{or} \quad x = \ln(2)$

Would you like to know more details or have any questions?

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Here are 5 questions that expand on this:

1. How would you solve a quadratic equation if it couldn't be factored easily?
2. What are some properties of logarithms that are useful in solving equations like this?
3. How would the equation change if it were $e^{2x} - 6e^x + 9 = 0$?
4. Can you solve $e^x + 2e^{-x} = 3$ for $x$?
5. What happens when we have negative solutions for $u$ in terms of $e^x$?

Tip: Remember, for solving quadratic equations, factoring is a quick method when possible, but using the quadratic formula works for any quadratic.