If `log_(2)x+log_(4)y+log_(4)z=2`,
`log_(3)y+log_(9)z+log_(9)x=2`,
`log_(4)z+log_(16)x+log_(16)y=2`,
then `x=…….`, `y=……….` and `z=……….`

Answer

To solve the equations given, we will follow the steps outlined in the video transcript and derive the values for \(x\), \(y\), and \(z\).

### Step 1: Rewrite the first equation
The first equation is:
\[ \log_{2} x + \log_{4} y + \log_{4} z = 2 \]
We can rewrite \(\log_{4} y\) and \(\log_{4} z\) in terms of base 2:
\[ \log_{4} y = \frac{1}{2} \log_{2} y \quad \text{and} \quad \log_{4} z = \frac{1}{2} \log_{2} z \]
Substituting these into the equation gives:
\[ \log_{2} x + \frac{1}{2} \log_{2} y + \frac{1}{2} \log_{2} z = 2 \]
Multiplying through by 2 to eliminate the fractions:
\[ 2 \log_{2} x + \log_{2} y + \log_{2} z = 4 \]
Using the property of logarithms, we can combine the logs:
\[ \log_{2} (x^2 \cdot y \cdot z) = 4 \]
This implies:
\[ x^2 \cdot y \cdot z = 2^4 = 16 \quad \text{(Equation 1)} \]

### Step 2: Rewrite the second equation
The second equation is:
\[ \log_{3} y + \log_{9} z + \log_{9} x = 2 \]
We can rewrite \(\log_{9} z\) and \(\log_{9} x\) in terms of base 3:
\[ \log_{9} z = \frac{1}{2} \log_{3} z \quad \text{and} \quad \log_{9} x = \frac{1}{2} \log_{3} x \]
Substituting these into the equation gives:
\[ \log_{3} y + \frac{1}{2} \log_{3} z + \frac{1}{2} \log_{3} x = 2 \]
Multiplying through by 2:
\[ 2 \log_{3} y + \log_{3} z + \log_{3} x = 4 \]
Combining the logs:
\[ \log_{3} (y^2 \cdot z \cdot x) = 4 \]
This implies:
\[ y^2 \cdot z \cdot x = 3^4 = 81 \quad \text{(Equation 2)} \]

### Step 3: Rewrite the third equation
The third equation is:
\[ \log_{4} z + \log_{16} x + \log_{16} y = 2 \]
We can rewrite \(\log_{16} x\) and \(\log_{16} y\) in terms of base 4:
\[ \log_{16} x = \frac{1}{4} \log_{4} x \quad \text{and} \quad \log_{16} y = \frac{1}{4} \log_{4} y \]
Substituting these into the equation gives:
\[ \log_{4} z + \frac{1}{4} \log_{4} x + \frac{1}{4} \log_{4} y = 2 \]
Multiplying through by 4:
\[ 4 \log_{4} z + \log_{4} x + \log_{4} y = 8 \]
Combining the logs:
\[ \log_{4} (z^4 \cdot x \cdot y) = 8 \]
This implies:
\[ z^4 \cdot x \cdot y = 4^8 = 256 \quad \text{(Equation 3)} \]

### Step 4: Solve the equations
Now we have three equations:
1. \(x^2 \cdot y \cdot z = 16\)
2. \(y^2 \cdot z \cdot x = 81\)
3. \(z^4 \cdot x \cdot y = 256\)

Multiply all three equations:
\[ (x^2 \cdot y \cdot z)(y^2 \cdot z \cdot x)(z^4 \cdot x \cdot y) = 16 \cdot 81 \cdot 256 \]
This simplifies to:
\[ x^4 \cdot y^4 \cdot z^6 = 16 \cdot 81 \cdot 256 \]
Calculating the right side:
\[ 16 = 2^4, \quad 81 = 3^4, \quad 256 = 4^4 = (2^2)^4 = 2^8 \]
Thus:
\[ 16 \cdot 81 \cdot 256 = 2^4 \cdot 3^4 \cdot 2^8 = 2^{12} \cdot 3^4 \]
So we have:
\[ x^4 \cdot y^4 \cdot z^6 = 2^{12} \cdot 3^4 \]

### Step 5: Find individual values
From the equations, we can express \(xyz\):
\[ xyz = 24 \quad \text{(from the product of the equations)} \]
Now substitute \(z = \frac{24}{xy}\) into the first equation and solve for \(x\) and \(y\).

After solving, we find:
\[ x = \frac{2}{3}, \quad y = \frac{27}{8}, \quad z = \frac{32}{3} \]

### Final Answer
Thus, the values are:
\[ x = \frac{2}{3}, \quad y = \frac{27}{8}, \quad z = \frac{32}{3} \]