If y = `a^((1)/( 1 - log_(a)x)) , z = a^((1)/(1 - log_(a)y))`, then x will be

Options

A. `a^((1)/(1 - log_(a)z))`
B. `log_(a^(z))`
C. `a^((1)/(1+ log_(a)z))`
D. none of these

Answer

To solve the problem, we start with the given equations:

1. \( y = a^{\frac{1}{1 - \log_a x}} \)
2. \( z = a^{\frac{1}{1 - \log_a y}} \)

We need to find the value of \( x \).

### Step 1: Take the logarithm of both sides of the first equation
Taking logarithm base \( a \) on both sides of the first equation:

\[ \log_a y = \frac{1}{1 - \log_a x} \]

### Step 2: Rearranging the first equation
From the equation \( \log_a y = \frac{1}{1 - \log_a x} \), we can express it as:

\[ 1 - \log_a x = \frac{1}{\log_a y} \]

### Step 3: Rearranging to isolate \( \log_a x \)
Rearranging gives us:

\[ \log_a x = 1 - \frac{1}{\log_a y} \]

### Step 4: Take the logarithm of both sides of the second equation
Now, we take the logarithm of both sides of the second equation:

\[ \log_a z = \frac{1}{1 - \log_a y} \]

### Step 5: Rearranging the second equation
From the equation \( \log_a z = \frac{1}{1 - \log_a y} \), we can express it as:

\[ 1 - \log_a y = \frac{1}{\log_a z} \]

### Step 6: Rearranging to isolate \( \log_a y \)
Rearranging gives us:

\[ \log_a y = 1 - \frac{1}{\log_a z} \]

### Step 7: Substitute \( \log_a y \) into the equation for \( \log_a x \)
Now we substitute the expression for \( \log_a y \) into the equation for \( \log_a x \):

\[ \log_a x = 1 - \frac{1}{1 - \frac{1}{\log_a z}} \]

### Step 8: Simplifying the expression for \( \log_a x \)
To simplify \( \log_a x \):

\[ \log_a x = 1 - \frac{\log_a z}{\log_a z - 1} \]

### Step 9: Finding \( x \)
Now, we can express \( x \) in terms of \( a \):

\[ x = a^{\log_a x} = a^{1 - \frac{\log_a z}{\log_a z - 1}} \]

### Step 10: Final expression for \( x \)
After simplifying, we find:

\[ x = \frac{a}{a^{\frac{\log_a z}{\log_a z - 1}}} \]

This gives us the final value of \( x \).