The value of `x^("log"_(x) a xx "log"_(a)y xx "log"_(y) z)` is

Options

A. x
B. y
C. z
D. a prime number

Explanation

To solve the expression \( x^{\log_{x} a \cdot \log_{a} y \cdot \log_{y} z} \), we will follow these steps:

### Step 1: Rewrite the logarithms using the change of base formula
We can express each logarithm in terms of natural logarithms (or common logarithms) using the change of base formula:
\[ \log_{b} a = \frac{\log a}{\log b} \]
Thus, we can rewrite the expression:
\[ \log_{x} a = \frac{\log a}{\log x}, \quad \log_{a} y = \frac{\log y}{\log a}, \quad \log_{y} z = \frac{\log z}{\log y} \]

### Step 2: Substitute the rewritten logarithms into the expression
Substituting these into the original expression gives:
\[ x^{\frac{\log a}{\log x} \cdot \frac{\log y}{\log a} \cdot \frac{\log z}{\log y}} \]

### Step 3: Simplify the expression
Notice that the \(\log a\) in the numerator of \(\log_{x} a\) and the denominator of \(\log_{a} y\) cancels out, as does the \(\log y\) in the numerator of \(\log_{a} y\) and the denominator of \(\log_{y} z\):
\[ x^{\frac{\log z}{\log x}} \]

### Step 4: Apply the property of exponents
Using the property of exponents, we can rewrite the expression:
\[ x^{\frac{\log z}{\log x}} = z^{\log_{x} x} = z^1 \]

### Step 5: Final result
Thus, we find that:
\[ x^{\log_{x} a \cdot \log_{a} y \cdot \log_{y} z} = z \]

### Conclusion
The value of the expression is \( z \).

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