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

Options

A. x + y + z
B. xyz
C. `(1)/(xyz) `
D. none of these

Answer

To solve the problem, we need to find the value of \( xy + yz + zx \) given the expressions for \( x \), \( y \), and \( z \):

1. **Given Expressions:**
- \( x = 1 + \log_a(bc) \)
- \( y = 1 + \log_b(ca) \)
- \( z = 1 + \log_c(ab) \)

2. **Using the Change of Base Formula:**
We can rewrite the logarithmic terms using the change of base formula:
\[ \log_a(bc) = \frac{\log(bc)}{\log(a)} = \frac{\log(b) + \log(c)}{\log(a)} \]
Similarly,
\[ \log_b(ca) = \frac{\log(c) + \log(a)}{\log(b)} \]
\[ \log_c(ab) = \frac{\log(a) + \log(b)}{\log(c)} \]

3. **Substituting Back:**
Now substituting these back into the expressions for \( x \), \( y \), and \( z \):
\[ x = 1 + \frac{\log(b) + \log(c)}{\log(a)} = \frac{\log(a) + \log(b) + \log(c)}{\log(a)} \]
\[ y = 1 + \frac{\log(c) + \log(a)}{\log(b)} = \frac{\log(a) + \log(b) + \log(c)}{\log(b)} \]
\[ z = 1 + \frac{\log(a) + \log(b)}{\log(c)} = \frac{\log(a) + \log(b) + \log(c)}{\log(c)} \]

4. **Finding \( xy + yz + zx \):**
Now we can compute \( xy + yz + zx \):
\[ xy = \left(\frac{\log(a) + \log(b) + \log(c)}{\log(a)}\right)\left(\frac{\log(a) + \log(b) + \log(c)}{\log(b)}\right) = \frac{(\log(a) + \log(b) + \log(c))^2}{\log(a) \log(b)} \]
\[ yz = \left(\frac{\log(a) + \log(b) + \log(c)}{\log(b)}\right)\left(\frac{\log(a) + \log(b) + \log(c)}{\log(c)}\right) = \frac{(\log(a) + \log(b) + \log(c))^2}{\log(b) \log(c)} \]
\[ zx = \left(\frac{\log(a) + \log(b) + \log(c)}{\log(c)}\right)\left(\frac{\log(a) + \log(b) + \log(c)}{\log(a)}\right) = \frac{(\log(a) + \log(b) + \log(c))^2}{\log(c) \log(a)} \]

5. **Combining the Results:**
Now, adding these three results together:
\[ xy + yz + zx = \frac{(\log(a) + \log(b) + \log(c))^2}{\log(a) \log(b)} + \frac{(\log(a) + \log(b) + \log(c))^2}{\log(b) \log(c)} + \frac{(\log(a) + \log(b) + \log(c))^2}{\log(c) \log(a)} \]

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

6. **Final Result:**
Since the terms inside the parentheses sum up to 1, we have:
\[ xy + yz + zx = (\log(a) + \log(b) + \log(c))^2 \]

**Final Answer:**
\[ xy + yz + zx = xyz \]