We are given the following system of equations:

$2^{x + y} = 125 \quad \text{and} \quad 2^{x - y} = 25$

We are asked to determine the value of $2^{\frac{y}{x}}$ in closed form.

Step 1: Expressing the equations with logarithms

Start by rewriting the given powers in terms of logarithms. First, express 125 and 25 as powers of 5:

$125 = 5^3 \quad \text{and} \quad 25 = 5^2$

Thus, we rewrite the system as:

$2^{x + y} = 5^3 \quad \text{and} \quad 2^{x - y} = 5^2$

Step 2: Taking logarithms on both sides

Take the logarithm of both sides of each equation. Using natural logarithms (or base 2 logs) is appropriate, but we proceed generally:

For the first equation: $x + y = \log_2(5^3) = 3\log_2(5)$

For the second equation: $x - y = \log_2(5^2) = 2\log_2(5)$

Step 3: Solving the system of linear equations

We now have the following system of linear equations:

$x + y = 3\log_2(5)$ $x - y = 2\log_2(5)$

Add the two equations to eliminate $y$:

$2x = 5\log_2(5)$

Thus, $x = \frac{5\log_2(5)}{2}$

Now, subtract the second equation from the first to eliminate $x$:

$2y = \log_2(5)$

Thus, $y = \frac{\log_2(5)}{2}$

Step 4: Finding $2^{\frac{y}{x}}$

We want to find $2^{\frac{y}{x}}$. From the expressions for $x$ and $y$, we have:

$\frac{y}{x} = \frac{\frac{\log_2(5)}{2}}{\frac{5\log_2(5)}{2}} = \frac{1}{5}$

Therefore: $2^{\frac{y}{x}} = 2^{\frac{1}{5}}$

Thus, the value of $2^{\frac{y}{x}}$ in closed form is $2^{\frac{1}{5}}$.

Would you like further details or clarifications?

Here are 5 related questions:

1. How would you solve this if the numbers were different powers of a different base, like 3 or 7?
2. What is the significance of taking logarithms in solving exponential equations?
3. Can you find a general method for solving equations of the form $a^{x+y} = m$ and $a^{x-y} = n$?
4. How does this approach change if the system involved more variables, such as $x+y+z$?
5. What other mathematical techniques can be used for solving nonlinear systems involving exponentials?

Tip: When solving exponential equations, expressing the constants as powers of the same base can greatly simplify the solution process.