To solve the equation $e^{2x} - 4e^x + 3 = 0$, let's proceed by substituting a variable to simplify it.

1. Let $u = e^x$. Then, $e^{2x} = u^2$.

   Substitute these into the equation: $u^2 - 4u + 3 = 0$

2. Now we have a quadratic equation in terms of $u$: $u^2 - 4u + 3 = 0$

3. Factor the quadratic equation: $(u - 1)(u - 3) = 0$

4. Set each factor equal to zero: $u - 1 = 0 \quad \Rightarrow \quad u = 1$ $u - 3 = 0 \quad \Rightarrow \quad u = 3$

5. Now, substitute back $u = e^x$ to solve for $x$:

   • For $u = 1$: $e^x = 1 \quad \Rightarrow \quad x = \ln(1) = 0$

   • For $u = 3$: $e^x = 3 \quad \Rightarrow \quad x = \ln(3)$

6. Calculate $\ln(3)$ to four decimal places: $x = \ln(3) \approx 1.0986$

Final answers:

• $x = 0$ (smaller value)
• $x \approx 1.0986$ (larger value)

Would you like further details on this solution, or do you have any questions?

---

Related Questions:

1. How would this solution change if the equation were $e^{2x} - 5e^x + 6 = 0$?
2. Can you solve similar equations where the exponent term is more complex, such as $e^{3x} - 7e^{x} + 10 = 0$?
3. How can we check if our solutions are correct by substituting them back into the original equation?
4. What are some applications of solving exponential equations like this in real-world contexts?
5. How can logarithmic properties help solve equations with different bases in exponentials?

Tip: When dealing with exponential equations, substitution can simplify the problem, making it easier to solve by turning it into a quadratic or other familiar form.