If a, b, c are positive real numbers, then
`a^("log"b-"log"c) xx b^("log"c-"log"a) xx c^("log"a - "log"b)`

Options

A. 0
B. 1
C. `-1`
D. none of these

Explanation

To solve the expression \( a^{(\log b - \log c)} \cdot b^{(\log c - \log a)} \cdot c^{(\log a - \log b)} \), we can use properties of logarithms and exponents. Here’s the step-by-step solution:

### Step 1: Apply the Property of Logarithms
We can use the property of logarithms that states:
\[ \log m - \log n = \log \left(\frac{m}{n}\right) \]
Using this property, we rewrite the expression:
\[ a^{(\log b - \log c)} = a^{\log\left(\frac{b}{c}\right)} \]
\[ b^{(\log c - \log a)} = b^{\log\left(\frac{c}{a}\right)} \]
\[ c^{(\log a - \log b)} = c^{\log\left(\frac{a}{b}\right)} \]

### Step 2: Rewrite the Entire Expression
Now we can rewrite the entire expression:
\[ a^{\log\left(\frac{b}{c}\right)} \cdot b^{\log\left(\frac{c}{a}\right)} \cdot c^{\log\left(\frac{a}{b}\right)} \]

### Step 3: Apply the Exponential Property
Using the property \( x^{\log_y z} = z^{\log_y x} \), we can rewrite each term:
\[ a^{\log\left(\frac{b}{c}\right)} = \left(\frac{b}{c}\right)^{\log a} \]
\[ b^{\log\left(\frac{c}{a}\right)} = \left(\frac{c}{a}\right)^{\log b} \]
\[ c^{\log\left(\frac{a}{b}\right)} = \left(\frac{a}{b}\right)^{\log c} \]

### Step 4: Combine the Terms
Now we can combine these terms:
\[ \left(\frac{b}{c}\right)^{\log a} \cdot \left(\frac{c}{a}\right)^{\log b} \cdot \left(\frac{a}{b}\right)^{\log c} \]

### Step 5: Simplify the Expression
This can be simplified as follows:
\[ = \frac{b^{\log a}}{c^{\log a}} \cdot \frac{c^{\log b}}{a^{\log b}} \cdot \frac{a^{\log c}}{b^{\log c}} \]
Now, we can rearrange and combine the fractions:
\[ = \frac{b^{\log a} \cdot c^{\log b} \cdot a^{\log c}}{c^{\log a} \cdot a^{\log b} \cdot b^{\log c}} \]

### Step 6: Observe the Cancellation
Notice that each base appears in both the numerator and the denominator:
- \(b^{\log a}\) in the numerator cancels with \(b^{\log c}\) in the denominator.
- \(c^{\log b}\) in the numerator cancels with \(c^{\log a}\) in the denominator.
- \(a^{\log c}\) in the numerator cancels with \(a^{\log b}\) in the denominator.

Thus, we are left with:
\[ = 1 \]

### Final Answer
The final answer is:
\[ \boxed{1} \]