If `x=log_(a)(bc),y=log_(b)(ca)" and "z=log_(c)(ab)`, then which of the following is equal to 1 ?

Options

A. `x+y+z`
B. `(1+x)^(-1)+(1+y)^(-1)+(1+z)^(-1)`
C. xyz
D. None of these

Answer

To solve the problem, we need to analyze the given logarithmic expressions and find which of the options is equal to 1.

Given:
- \( x = \log_a(bc) \)
- \( y = \log_b(ca) \)
- \( z = \log_c(ab) \)

We want to find which of the following expressions is equal to 1.

### Step 1: Rewrite the logarithmic expressions
Using the property of logarithms, we can rewrite each expression:
- \( x = \log_a(bc) = \log_a(b) + \log_a(c) \)
- \( y = \log_b(ca) = \log_b(c) + \log_b(a) \)
- \( z = \log_c(ab) = \log_c(a) + \log_c(b) \)

### Step 2: Analyze the expression \( \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} \)
We will check if the expression \( \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} \) equals 1.

### Step 3: Substitute the values of \( x, y, z \)
Substituting the values of \( x, y, z \):
\[ \frac{1}{1 + \log_a(bc)} + \frac{1}{1 + \log_b(ca)} + \frac{1}{1 + \log_c(ab)} \]

### Step 4: Use the change of base formula
Using the change of base formula, we can express these logarithms in terms of a common base (let's use base 10 for simplicity):
\[ \log_a(b) = \frac{\log(b)}{\log(a)}, \quad \log_b(c) = \frac{\log(c)}{\log(b)}, \quad \log_c(a) = \frac{\log(a)}{\log(c)} \]
Now, we can express \( x, y, z \) in terms of base 10 logarithms:
- \( x = \frac{\log(b) + \log(c)}{\log(a)} \)
- \( y = \frac{\log(c) + \log(a)}{\log(b)} \)
- \( z = \frac{\log(a) + \log(b)}{\log(c)} \)

### Step 5: Simplify the expression
Now we can simplify the expression:
\[ \frac{1}{1 + x} = \frac{\log(a)}{\log(a) + \log(b) + \log(c)}, \quad \frac{1}{1 + y} = \frac{\log(b)}{\log(a) + \log(b) + \log(c)}, \quad \frac{1}{1 + z} = \frac{\log(c)}{\log(a) + \log(b) + \log(c)} \]
Adding these fractions together:
\[ \frac{\log(a) + \log(b) + \log(c)}{\log(a) + \log(b) + \log(c)} = 1 \]

### Conclusion
Thus, we find that:
\[ \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1 \]
Therefore, the expression that equals 1 is:
\[ \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1 \]

### Final Answer
The option that is equal to 1 is:
\[ \text{Option 2: } \frac{1}{1 + x} + \frac{1}{1 + y} + \frac{1}{1 + z} = 1 \]