The given equation is:

$e^{2x} - 6e^x - 40 = 0$

This equation involves exponential functions, and we can simplify it by substituting $y = e^x$. Thus, $e^{2x} = (e^x)^2 = y^2$, and the equation becomes:

$y^2 - 6y - 40 = 0$

This is now a quadratic equation in terms of $y$. To solve it, we can use the quadratic formula:

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

For the quadratic equation $y^2 - 6y - 40 = 0$, we have:

• $a = 1$
• $b = -6$
• $c = -40$

Substituting these into the quadratic formula:

$y = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(-40)}}{2(1)}$ $y = \frac{6 \pm \sqrt{36 + 160}}{2}$ $y = \frac{6 \pm \sqrt{196}}{2}$ $y = \frac{6 \pm 14}{2}$

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

$y = \frac{6 + 14}{2} = 10 \quad \text{or} \quad y = \frac{6 - 14}{2} = -4$

Now, recall that $y = e^x$. Therefore, we have two cases:

1. $e^x = 10$
2. $e^x = -4$

Since the exponential function $e^x$ is always positive for any real value of $x$, the second case, $e^x = -4$, is not possible. Hence, we only consider the first case:

$e^x = 10$

Taking the natural logarithm of both sides:

$x = \ln(10)$

Thus, the value of $x$ that satisfies the equation is:

$x = \ln(10)$

This is the value you should enter.

Would you like more details or have any questions about the solution?

Further questions:

1. How does the exponential function behave for large values of $x$?
2. What are the properties of the natural logarithm function?
3. Can you solve other types of exponential equations similarly?
4. How would you solve a quadratic equation that doesn't factor easily?
5. Why is the solution $e^x = -4$ not valid?

Tip: When solving exponential equations, always check if negative solutions for $e^x$ make sense, as they are never possible.