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We are given the following two equations:

1. \( 2^{x+y} = 125 \)
2. \( 2^{x-y} = 25 \)

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

### Step 1: Express the equations in logarithmic form
We know that \( 125 = 5^3 \) and \( 25 = 5^2 \), so we can rewrite the equations as follows:

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

### Step 2: Take the logarithm on both sides
Taking the natural logarithm (or log base 2 for simplicity) on both sides of both equations, we get:

\[
x + y = \log_2 125 \quad \text{and} \quad x - y = \log_2 25
\]

### Step 3: Solve the system of equations
We now have the system of linear equations:

\[
x + y = \log_2 125 \quad (1)
\]
\[
x - y = \log_2 25 \quad (2)
\]

Add equations (1) and (2) to eliminate \(y\):

\[
(x + y) + (x - y) = \log_2 125 + \log_2 25
\]
\[
2x = \log_2 (125 \times 25) = \log_2 3125
\]

Now, solve for \(x\):

\[
x = \frac{1}{2} \log_2 3125
\]

Since \( 3125 = 5^5 \), we have:

\[
x = \frac{1}{2} \log_2 (5^5) = \frac{1}{2} \times 5 \log_2 5 = \frac{5 \log_2 5}{2}
\]

### Step 4: Solve for \(y\)
Now subtract equation (2) from equation (1) to eliminate \(x\):

\[
(x + y) - (x - y) = \log_2 125 - \log_2 25
\]
\[
2y = \log_2 \left(\frac{125}{25}\right) = \log_2 5
\]

So:

\[
y = \frac{1}{2} \log_2 5
\]

### Step 5: Find \( 2^{\frac{y}{x}} \)
Now we need to compute \( 2^{\frac{y}{x}} \). Using the values we found for \(y\) and \(x\):

\[
\frac{y}{x} = \frac{\frac{1}{2} \log_2 5}{\frac{5}{2} \log_2 5} = \frac{1}{5}
\]

Thus:

\[
2^{\frac{y}{x}} = 2^{\frac{1}{5}}
\]

### Final Answer:
The value of \( 2^{\frac{y}{x}} \) is \( 2^{\frac{1}{5}} \), which is the fifth root of 2.

Would you like any additional clarifications or details?

Here are some related questions to explore:
1. How can we use logarithms to solve other types of exponential equations?
2. What are the properties of logarithmic functions that help solve exponential systems?
3. What other methods can be used to solve systems of nonlinear equations like this?
4. How would the solution change if the exponents were fractions instead of integers?
5. Can this method be applied to exponential equations involving different bases, like base 3?

**Tip:** Always check if numbers can be rewritten as powers of smaller primes (like 2, 3, or 5) to simplify exponential equations.

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 exponential equations 2^(x+y) = 125 and 2^(x-y) = 25 and determine the value of 2^(y/x). This detailed step-by-step solution explores logarithmic methods and properties of exponents. Suitable for advanced high school students and early university levels.

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