If `log_(a)bc = x, log_(b)ca = y and log_(c)ab = z`, then the value of `(1)/(x +1) + (1)/(y +1) + (1)/(z+1)` is equal to

Options

A. 0
B. 1
C. 2
D. 3

Answer

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

1. \( \log_a(bc) = x \)
2. \( \log_b(ca) = y \)
3. \( \log_c(ab) = z \)

We need to find the value of:

\[ \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} \]

### Step 1: Rewrite the logarithmic expressions

Using the change of base formula, we can express \( x \), \( y \), and \( z \) as follows:

\[ x = \frac{\log(bc)}{\log(a)} = \frac{\log(b) + \log(c)}{\log(a)} \]

\[ y = \frac{\log(ca)}{\log(b)} = \frac{\log(c) + \log(a)}{\log(b)} \]

\[ z = \frac{\log(ab)}{\log(c)} = \frac{\log(a) + \log(b)}{\log(c)} \]

### Step 2: Substitute into the expression

Now we substitute \( x \), \( y \), and \( z \) into the expression we need to evaluate:

\[ \frac{1}{x + 1} = \frac{1}{\frac{\log(b) + \log(c)}{\log(a)} + 1} = \frac{\log(a)}{\log(b) + \log(c) + \log(a)} \]

\[ \frac{1}{y + 1} = \frac{1}{\frac{\log(c) + \log(a)}{\log(b)} + 1} = \frac{\log(b)}{\log(c) + \log(a) + \log(b)} \]

\[ \frac{1}{z + 1} = \frac{1}{\frac{\log(a) + \log(b)}{\log(c)} + 1} = \frac{\log(c)}{\log(a) + \log(b) + \log(c)} \]

### Step 3: Combine the fractions

Now we can add these fractions together:

\[ \frac{\log(a)}{\log(a) + \log(b) + \log(c)} + \frac{\log(b)}{\log(a) + \log(b) + \log(c)} + \frac{\log(c)}{\log(a) + \log(b) + \log(c)} \]

Since all three fractions have the same denominator, we can combine them:

\[ = \frac{\log(a) + \log(b) + \log(c)}{\log(a) + \log(b) + \log(c)} = 1 \]

### Final Result

Thus, the value of \( \frac{1}{x + 1} + \frac{1}{y + 1} + \frac{1}{z + 1} \) is:

\[ \boxed{1} \]