. If `1, log_y x, log_z y, -15 log_x z` are in AP, then

Options

A. `x=z^(3)`
B. `x=y^(-1)`
C. `y=z^(-3)`
D. `y=z^(3)`

Explanation

To solve the problem where \( 1, \log_y x, \log_z y, -15 \log_x z \) are in Arithmetic Progression (AP), we can follow these steps:

### Step 1: Understanding the AP condition
For four terms \( a, b, c, d \) to be in AP, the condition is:
\[ 2b = a + c \]
In our case:
- \( a = 1 \)
- \( b = \log_y x \)
- \( c = \log_z y \)
- \( d = -15 \log_x z \)

### Step 2: Setting up the equation
Using the AP condition, we can write:
\[ 2 \log_y x = 1 + \log_z y \]

### Step 3: Expressing logarithms in terms of a common base
Using the change of base formula, we can express the logarithms:
\[ \log_y x = \frac{\log x}{\log y}, \quad \log_z y = \frac{\log y}{\log z}, \quad \log_x z = \frac{\log z}{\log x} \]
Substituting these into our equation gives:
\[ 2 \frac{\log x}{\log y} = 1 + \frac{\log y}{\log z} \]

### Step 4: Cross-multiplying to eliminate fractions
Cross-multiplying gives:
\[ 2 \log x \cdot \log z = \log y \cdot (\log y + \log z) \]
This simplifies to:
\[ 2 \log x \cdot \log z = \log^2 y + \log y \cdot \log z \]

### Step 5: Rearranging the equation
Rearranging gives:
\[ \log^2 y + \log y \cdot \log z - 2 \log x \cdot \log z = 0 \]

### Step 6: Treating it as a quadratic in \( \log y \)
This is a quadratic equation in \( \log y \):
\[ \log^2 y + \log y \cdot \log z - 2 \log x \cdot \log z = 0 \]
Let \( u = \log y \). The equation becomes:
\[ u^2 + u \log z - 2 \log x \cdot \log z = 0 \]

### Step 7: Applying the quadratic formula
Using the quadratic formula \( u = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
\[ u = \frac{-\log z \pm \sqrt{(\log z)^2 + 8 \log x \log z}}{2} \]

### Step 8: Finding the values of \( y \)
Substituting back \( u = \log y \):
\[ \log y = \frac{-\log z \pm \sqrt{(\log z)^2 + 8 \log x \log z}}{2} \]
From this, we can find \( y \) as:
\[ y = 10^{\left(\frac{-\log z \pm \sqrt{(\log z)^2 + 8 \log x \log z}}{2}\right)} \]

### Step 9: Finding \( z \) in terms of \( x \)
Using the relationship \( -15 \log_x z \):
\[ \log_x z = \frac{\log z}{\log x} \]
We substitute this back into the original equation to find the relationship between \( x \), \( y \), and \( z \).

### Conclusion
After solving the quadratic and substituting back, we can find the relationships among \( x, y, z \) and verify which options are correct based on the derived equations.