The given equation is:

$8e^{-x} - e^x = 2$

Let's solve this step by step.

Step 1: Substitute $e^{-x}$

We can rewrite $e^{-x}$ as $\frac{1}{e^x}$, so the equation becomes:

$8\frac{1}{e^x} - e^x = 2$

Step 2: Multiply through by $e^x$

To eliminate the fraction, multiply both sides of the equation by $e^x$:

$8 - e^{2x} = 2e^x$

Step 3: Rearrange the equation

Move all terms to one side to form a quadratic equation in terms of $e^x$:

$e^{2x} + 2e^x - 8 = 0$

Step 4: Use substitution

Let $u = e^x$. The equation now becomes a quadratic in $u$:

$u^2 + 2u - 8 = 0$

Step 5: Solve the quadratic equation

We can solve this using the quadratic formula. The quadratic formula is:

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

For the equation $u^2 + 2u - 8 = 0$, $a = 1$, $b = 2$, and $c = -8$. Plugging these into the formula:

$u = \frac{-2 \pm \sqrt{2^2 - 4(1)(-8)}}{2(1)}$ $u = \frac{-2 \pm \sqrt{4 + 32}}{2}$ $u = \frac{-2 \pm \sqrt{36}}{2}$ $u = \frac{-2 \pm 6}{2}$

So, the two solutions for $u$ are:

$u = \frac{-2 + 6}{2} = 2 \quad \text{or} \quad u = \frac{-2 - 6}{2} = -4$

Step 6: Back-substitute $u = e^x$

Now, recall that $u = e^x$, so:

1. $e^x = 2$ ⟹ $x = \ln(2)$
2. $e^x = -4$ is not valid because $e^x$ is always positive.

Final Answer:

The only valid solution is:

$x = \ln(2)$

Would you like more details or explanations?

Here are 5 related questions to explore further:

1. How can you apply logarithmic rules to simplify other exponential equations?
2. What happens when you have multiple valid solutions in quadratic exponential equations?
3. How can we solve similar equations with different bases, like $e^x$ and $2^x$?
4. What real-world applications involve solving exponential equations like this one?
5. How can we use graphical methods to solve exponential equations?

Tip: Always check the validity of solutions when working with exponential equations, especially for constraints like positivity.