The given equation is:

$e^{2x} - 4e^x = 5$

We can solve for $x$ by making a substitution. Let:

$y = e^x$

Now, rewrite the equation in terms of $y$:

$y^2 - 4y = 5$

This is a quadratic equation. Rearrange it:

$y^2 - 4y - 5 = 0$

Now, we solve this quadratic equation. We can use the quadratic formula:

$y = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$

where $a = 1$, $b = -4$, and $c = -5$. Substituting these values into the formula:

$y = \frac{-(-4) \pm \sqrt{(-4)^2 - 4(1)(-5)}}{2(1)}$ $y = \frac{4 \pm \sqrt{16 + 20}}{2}$ $y = \frac{4 \pm \sqrt{36}}{2}$ $y = \frac{4 \pm 6}{2}$

Thus, we have two possible solutions for $y$:

1. $y = \frac{4 + 6}{2} = \frac{10}{2} = 5$
2. $y = \frac{4 - 6}{2} = \frac{-2}{2} = -1$

Recall that $y = e^x$. Therefore, we now have two cases to consider:

Case 1: $e^x = 5$

Taking the natural logarithm (ln) of both sides:

$x = \ln(5)$

Case 2: $e^x = -1$

Since $e^x$ is always positive for any real $x$, there is no solution for this case.

Thus, the only valid solution is:

$x = \ln(5)$

Let me know if you'd like further clarification or have any other questions!

Related Questions:

1. What is the natural logarithm of 5?
2. How do you solve a quadratic equation using the quadratic formula?
3. What are the properties of the exponential function $e^x$?
4. Can you explain the concept of logarithms and their use in solving exponential equations?
5. How can this method be applied to other similar exponential equations?

Tip: When solving equations involving exponentials, always look for substitution opportunities to simplify the equation, as it can often turn an exponential equation into a quadratic one!