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Let's solve the given problem step-by-step.

We are given two equations:
1. \(2^{x + y} = 125\)
2. \(2^{x - y} = 25\)

We are asked to determine the value of \(2^{\frac{y}{x}}\) in **closed form** without using a calculator.

---

### Step 1: Rewrite the constants in terms of powers of 2
We know:
- \(125 = 5^3\)
- \(25 = 5^2\)

Thus, the equations become:
\[
2^{x + y} = 5^3 \quad \text{and} \quad 2^{x - y} = 5^2.
\]

---

### Step 2: Take logarithms to make the exponents manageable
Using the **base-2 logarithm** (denoted by \(\log_2\)), we take the log of both sides for each equation.

For the first equation:
\[
\log_2 \left(2^{x + y}\right) = \log_2 \left(5^3\right) \implies x + y = 3 \cdot \log_2(5).
\]

For the second equation:
\[
\log_2 \left(2^{x - y}\right) = \log_2 \left(5^2\right) \implies x - y = 2 \cdot \log_2(5).
\]

---

### Step 3: Solve the system of linear equations
We have:
\[
x + y = 3 \log_2(5), \quad x - y = 2 \log_2(5).
\]

Add the two equations:
\[
2x = 5 \log_2(5) \implies x = \frac{5}{2} \log_2(5).
\]

Now subtract the second equation from the first:
\[
2y = \log_2(5) \implies y = \frac{1}{2} \log_2(5).
\]

---

### Step 4: Find \( \frac{y}{x} \)
\[
\frac{y}{x} = \frac{\frac{1}{2} \log_2(5)}{\frac{5}{2} \log_2(5)} = \frac{1/2}{5/2} = \frac{1}{5}.
\]

---

### Step 5: Compute \( 2^{\frac{y}{x}} \)
\[
2^{\frac{y}{x}} = 2^{\frac{1}{5}}.
\]

---

### Final Answer
The value of \(2^{\frac{y}{x}}\) in closed form is:
\[
\boxed{2^{\frac{1}{5}}}.
\]

---

Do you have any questions or want further clarification?  
Here are 5 related questions to explore further:

1. What are the general properties of exponential functions with fractional exponents?
2. How can logarithmic identities simplify solving equations involving exponents?
3. What is the relationship between different logarithmic bases (e.g., \(\log_2\) and \(\log_{10}\))?
4. How do you solve more complex systems involving logarithmic equations?
5. Can this approach be extended to solve non-linear exponential equations?

**Tip:** When handling exponential equations, it often helps to use logarithms, especially when unknowns appear in the exponent.

Suppose 2^(x + y) = 125 and 2^(x - y) = 25. Without using a calculator, determine in closed form the value of 2^(y/x).

Learn how to solve the system of exponential equations 2^(x + y) = 125 and 2^(x - y) = 25. This problem covers the application of exponents and logarithmic identities to determine the closed form value of 2^(y/x). Suitable for students in Grades 10-12.

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