Solve for `x, y , z`.
`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`

Options

A. `(175)/(12)`
B. `(349)/(24)`
C. `(353)/(24)`
D. `(112)/(3)`

Answer

To solve the equations given in the problem, we will follow a systematic approach using logarithmic properties.

### Given Equations:
1. \( \log_2 x + \log_4 y + \log_4 z = 2 \)
2. \( \log_3 y + \log_9 z + \log_9 x = 2 \)
3. \( \log_4 z + \log_{16} x + \log_{16} y = 2 \)

### Step 1: Simplifying the First Equation
We start with the first equation:

\[ \log_2 x + \log_4 y + \log_4 z = 2 \]

We can convert the logarithms with base 4 to 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, we get:

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

Multiplying the entire equation by 2 to eliminate the fractions:

\[ 2 \log_2 x + \log_2 y + \log_2 z = 4 \]

Using the property of logarithms that states \( \log_a b + \log_a c = \log_a (bc) \):

\[ \log_2 (x^2) + \log_2 (y) + \log_2 (z) = 4 \]

This simplifies to:

\[ \log_2 (x^2 y z) = 4 \]

Exponentiating both sides gives:

\[ x^2 y z = 16 \quad \text{(Equation 1)} \]

### Step 2: Simplifying the Second Equation
Now we move to the second equation:

\[ \log_3 y + \log_9 z + \log_9 x = 2 \]

Converting the logarithms with base 9 to 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 the entire equation by 2:

\[ 2 \log_3 y + \log_3 z + \log_3 x = 4 \]

This simplifies to:

\[ \log_3 (y^2) + \log_3 (z) + \log_3 (x) = 4 \]

Thus:

\[ \log_3 (y^2 x z) = 4 \]

Exponentiating both sides gives:

\[ y^2 x z = 81 \quad \text{(Equation 2)} \]

### Step 3: Simplifying the Third Equation
Now we simplify the third equation:

\[ \log_4 z + \log_{16} x + \log_{16} y = 2 \]

Converting the logarithms with base 16 to base 4:

\[ \log_{16} x = \frac{1}{2} \log_4 x \quad \text{and} \quad \log_{16} y = \frac{1}{2} \log_4 y \]

Substituting gives:

\[ \log_4 z + \frac{1}{2} \log_4 x + \frac{1}{2} \log_4 y = 2 \]

Multiplying the entire equation by 2:

\[ 2 \log_4 z + \log_4 x + \log_4 y = 4 \]

This simplifies to:

\[ \log_4 (z^2) + \log_4 (x) + \log_4 (y) = 4 \]

Thus:

\[ \log_4 (z^2 x y) = 4 \]

Exponentiating both sides gives:

\[ z^2 x y = 256 \quad \text{(Equation 3)} \]

### Step 4: Solving the System of Equations
Now we have three equations:

1. \( x^2 y z = 16 \)
2. \( y^2 x z = 81 \)
3. \( z^2 x y = 256 \)

We can express \( z \) from Equation 1:

\[ z = \frac{16}{x^2 y} \]

Substituting \( z \) into Equation 2:

\[ y^2 x \left(\frac{16}{x^2 y}\right) = 81 \]

This simplifies to:

\[ \frac{16y}{x} = 81 \implies y = \frac{81x}{16} \]

Now substituting \( y \) back into the expression for \( z \):

\[ z = \frac{16}{x^2 \left(\frac{81x}{16}\right)} = \frac{256}{81x^3} \]

### Step 5: Substitute into Equation 3
Now substitute \( y \) and \( z \) into Equation 3:

\[ \left(\frac{256}{81x^3}\right)^2 x \left(\frac{81x}{16}\right) = 256 \]

This simplifies to:

\[ \frac{65536}{6561 x^6} \cdot x \cdot \frac{81x}{16} = 256 \]

After solving this equation, we find:

\[ x = \frac{2}{3} \]

### Step 6: Finding \( y \) and \( z \)
Now substituting \( x \) back into the expressions for \( y \) and \( z \):

\[ y = \frac{81 \cdot \frac{2}{3}}{16} = \frac{27}{8} \]

\[ z = \frac{256}{81 \cdot \left(\frac{2}{3}\right)^3} = \frac{32}{3} \]

### Final Values
Thus, the values of \( x, y, z \) are:

\[ x = \frac{2}{3}, \quad y = \frac{27}{8}, \quad z = \frac{32}{3} \]