The problem you're working with asks to solve the equation:

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

in the real numbers $\mathbb{R}$, and the answer should be given in the form of a set of solutions.

Solution:

Since the equation involves exponential functions with the same base $e$, we can equate the exponents directly:

$-2x - 5 = x$

Now, let's solve for $x$:

1. Move all terms involving $x$ to one side: $-2x - x = 5$ $-3x = 5$

2. Solve for $x$ by dividing both sides by $-3$: $x = \frac{5}{-3} = -\frac{5}{3}$

Final Answer:

The solution set is:

$\left\{-\frac{5}{3}\right\}$

Would you like more details or have any questions?

Here are 5 related questions for further exploration:

1. How do we solve exponential equations with different bases?
2. What happens if we add logarithms to both sides of an exponential equation?
3. Can exponential equations always be solved algebraically?
4. How would the solution change if the equation was $e^{-2x - 5} = e^{2x}$?
5. How does the exponential function behave as $x$ approaches infinity?

Tip: When solving exponential equations, it’s helpful to simplify by isolating the exponents or converting to logarithmic form when needed.