Consider equations ` x^(log_(y)x) = 2 and y^(log_(x)y) = 16`.
The value of x is

Options

A. `2^(root(3)2)`
B. ` 2^(root(3)4)`
C. ` 2^(root(3)64)`
D. ` 2 root(3)256)`

Explanation

To solve the equations \( x^{\log_y x} = 2 \) and \( y^{\log_x y} = 16 \), we will follow these steps:

### Step 1: Define Variables
Let \( t = \log_y x \). Then, we can express \( x \) in terms of \( y \):
\[ x = y^t \]

### Step 2: Substitute into the First Equation
Substituting \( x = y^t \) into the first equation \( x^{\log_y x} = 2 \):
\[ (y^t)^{\log_y (y^t)} = 2 \]
Using the property of logarithms, \( \log_y (y^t) = t \):
\[ (y^t)^t = 2 \]
This simplifies to:
\[ y^{t^2} = 2 \]

### Step 3: Express \( y \) in Terms of \( t \)
From the equation \( y^{t^2} = 2 \), we can express \( y \):
\[ y = 2^{\frac{1}{t^2}} \]

### Step 4: Substitute into the Second Equation
Now, substitute \( y = 2^{\frac{1}{t^2}} \) into the second equation \( y^{\log_x y} = 16 \):
\[ (2^{\frac{1}{t^2}})^{\log_x (2^{\frac{1}{t^2}})} = 16 \]
Using the property of logarithms, \( \log_x (2^{\frac{1}{t^2}}) = \frac{1}{t^2} \log_x 2 \). We can express \( \log_x 2 \) in terms of \( t \):
\[ \log_x 2 = \frac{\log_y 2}{\log_y x} = \frac{\log_y 2}{t} \]
Thus, we have:
\[ (2^{\frac{1}{t^2}})^{\frac{1}{t^2} \log_x 2} = 16 \]
This simplifies to:
\[ 2^{\frac{1}{t^2} \cdot \frac{1}{t^2} \log_y 2} = 16 \]
Since \( 16 = 2^4 \), we equate the exponents:
\[ \frac{1}{t^4} \log_y 2 = 4 \]

### Step 5: Solve for \( t \)
Rearranging gives:
\[ \log_y 2 = 4t^4 \]
Using \( \log_y 2 = \frac{1}{\log_2 y} \):
\[ \frac{1}{\log_2 y} = 4t^4 \]
Thus:
\[ \log_2 y = \frac{1}{4t^4} \]

### Step 6: Substitute Back to Find \( x \)
Now we have expressions for both \( x \) and \( y \) in terms of \( t \). Substitute \( y = 2^{\frac{1}{t^2}} \) back into the expression for \( x \):
\[ x = y^t = \left(2^{\frac{1}{t^2}}\right)^t = 2^{\frac{t}{t^2}} = 2^{\frac{1}{t}} \]

### Step 7: Find \( t \)
From \( 4t^4 = \log_y 2 \), we can substitute \( y \) back and solve for \( t \):
\[ 4t^4 = \frac{1}{4t^4} \implies 16t^8 = 1 \implies t^8 = \frac{1}{16} \implies t = \frac{1}{2} \]

### Step 8: Substitute \( t \) Back to Find \( x \)
Now substituting \( t = \frac{1}{2} \) back into the expression for \( x \):
\[ x = 2^{\frac{1}{t}} = 2^{\frac{1}{\frac{1}{2}}} = 2^2 = 4 \]

### Final Answer
Thus, the value of \( x \) is:
\[ \boxed{4} \]