https://file.aiquickdraw.com/mathbot/file/17290931533320oz1gdrv.jpg

The system of equations given in the image is:

\[
\begin{cases}
x^{\log_2 y} + y^{\log_2 x} = 16, \\
\log_2 x - \log_2 y = 2.
\end{cases}
\]

### Step-by-Step Solution:

#### Step 1: Simplify the second equation
From the second equation, we have:
\[
\log_2 x - \log_2 y = 2.
\]
Using the logarithmic property \(\log_b \frac{a}{c} = \log_b a - \log_b c\), we can rewrite this as:
\[
\log_2 \frac{x}{y} = 2.
\]
Converting this logarithmic equation to an exponential form:
\[
\frac{x}{y} = 2^2 = 4.
\]
So, \(x = 4y\).

#### Step 2: Substitute into the first equation
Now that we know \(x = 4y\), substitute this into the first equation \(x^{\log_2 y} + y^{\log_2 x} = 16\).

Substitute \(x = 4y\) into both terms:
- The first term is \(x^{\log_2 y} = (4y)^{\log_2 y}\).
- The second term is \(y^{\log_2 x} = y^{\log_2 (4y)}\).

##### Simplify these terms:
- \( (4y)^{\log_2 y} = 4^{\log_2 y} \cdot y^{\log_2 y} \). We know that \(y^{\log_2 y} = y^{\log_2 y}\), and \(4^{\log_2 y} = (2^2)^{\log_2 y} = y^2\).
Thus, \( (4y)^{\log_2 y} = y^2 \cdot y = y^3 \).

- For the second term \( y^{\log_2 (4y)} \), we use the logarithmic property: 
\[
\log_2 (4y) = \log_2 4 + \log_2 y = 2 + \log_2 y.
\]
Thus, 
\[
y^{\log_2 (4y)} = y^{2 + \log_2 y} = y^2 \cdot y^{\log_2 y} = y^2 \cdot y = y^3.
\]

#### Step 3: Solve the equation
After substitution, the first equation becomes:
\[
y^3 + y^3 = 16,
\]
or
\[
2y^3 = 16.
\]
Dividing both sides by 2:
\[
y^3 = 8.
\]
Taking the cube root of both sides:
\[
y = 2.
\]

#### Step 4: Find \(x\)
Since \(x = 4y\), we substitute \(y = 2\):
\[
x = 4 \times 2 = 8.
\]

### Final Solution:
The values of \(x\) and \(y\) are:
\[
x = 8, \quad y = 2.
\]

---

Would you like any further details or have any questions? Here are 5 related questions you might consider:

1. What happens if the logarithmic bases in the system are different?
2. How would the solution change if the equation had a different constant on the right-hand side, say 32 instead of 16?
3. What are other properties of logarithms that are useful in solving exponential-logarithmic systems?
4. Can the logarithmic equations be solved graphically, and if so, how would that method work?
5. How do we generalize this method to solve more complex logarithmic systems?

**Tip:** Remember that converting between logarithmic and exponential forms is key when solving logarithmic equations.

Solve the system of equations: 

x^(log₂y) + y^(log₂x) = 16, 
log₂x - log₂y = 2.

A step-by-step solution to solve the system of logarithmic equations where x^(log₂y) + y^(log₂x) = 16, and log₂x - log₂y = 2. This problem explores logarithmic properties, exponential rules, and algebraic manipulations, ideal for high school students in Grades 10-12.

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