If `x=1+log_(a) bc, y=1+log_(b) ca, z=1+log_(c) ab`, then `(xyz)/(xy+yz+zx)` is equal to

Options

A. 2
B. 1
C. 4
D. 6

Answer

To solve the problem, we need to find the value of \(\frac{xyz}{xy + yz + zx}\) where:

\[ x = 1 + \log_a(bc), \quad y = 1 + \log_b(ca), \quad z = 1 + \log_c(ab) \]

### Step 1: Rewrite \(x\), \(y\), and \(z\)

We can express \(x\), \(y\), and \(z\) using properties of logarithms:

\[ x = 1 + \log_a(bc) = \log_a(a) + \log_a(bc) = \log_a(a \cdot bc) = \log_a(abc) \]

\[ y = 1 + \log_b(ca) = \log_b(b) + \log_b(ca) = \log_b(b \cdot ca) = \log_b(abc) \]

\[ z = 1 + \log_c(ab) = \log_c(c) + \log_c(ab) = \log_c(c \cdot ab) = \log_c(abc) \]

### Step 2: Calculate \(xyz\)

Now, we can find \(xyz\):

\[ xyz = \log_a(abc) \cdot \log_b(abc) \cdot \log_c(abc) \]

Using the change of base formula, we can rewrite each logarithm:

\[ xyz = \frac{\log(abc)}{\log(a)} \cdot \frac{\log(abc)}{\log(b)} \cdot \frac{\log(abc)}{\log(c)} = \frac{(\log(abc))^3}{\log(a) \log(b) \log(c)} \]

### Step 3: Calculate \(xy + yz + zx\)

Next, we calculate \(xy + yz + zx\):

\[ xy = \log_a(abc) \cdot \log_b(abc) = \frac{\log(abc)}{\log(a)} \cdot \frac{\log(abc)}{\log(b)} = \frac{(\log(abc))^2}{\log(a) \log(b)} \]

\[ yz = \log_b(abc) \cdot \log_c(abc) = \frac{\log(abc)}{\log(b)} \cdot \frac{\log(abc)}{\log(c)} = \frac{(\log(abc))^2}{\log(b) \log(c)} \]

\[ zx = \log_c(abc) \cdot \log_a(abc) = \frac{\log(abc)}{\log(c)} \cdot \frac{\log(abc)}{\log(a)} = \frac{(\log(abc))^2}{\log(c) \log(a)} \]

Now, summing these:

\[ xy + yz + zx = \frac{(\log(abc))^2}{\log(a) \log(b)} + \frac{(\log(abc))^2}{\log(b) \log(c)} + \frac{(\log(abc))^2}{\log(c) \log(a)} \]

Taking \((\log(abc))^2\) as a common factor:

\[ xy + yz + zx = (\log(abc))^2 \left( \frac{1}{\log(a) \log(b)} + \frac{1}{\log(b) \log(c)} + \frac{1}{\log(c) \log(a)} \right) \]

### Step 4: Calculate \(\frac{xyz}{xy + yz + zx}\)

Now we can find:

\[ \frac{xyz}{xy + yz + zx} = \frac{\frac{(\log(abc))^3}{\log(a) \log(b) \log(c)}}{(\log(abc))^2 \left( \frac{1}{\log(a) \log(b)} + \frac{1}{\log(b) \log(c)} + \frac{1}{\log(c) \log(a)} \right)} \]

This simplifies to:

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

### Step 5: Final Simplification

The denominator can be rewritten as:

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

Thus, we have:

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

After simplification, we find that this expression equals \(1\).

### Conclusion

Thus, the final answer is:

\[ \frac{xyz}{xy + yz + zx} = 1 \]