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

Options

A. 1
B. 0
C. abc
D. none of these

Answer

To solve the problem, we need to find the value of \( ab + bc + ca \) given the expressions for \( a \), \( b \), and \( c \).

### Step 1: Write down the expressions for \( a \), \( b \), and \( c \)

Given:
\[ a = 1 + \log_x(yz) \]
\[ b = 1 + \log_y(zx) \]
\[ c = 1 + \log_z(xy) \]

### Step 2: Convert the logarithmic expressions

Using the change of base formula, we can express the logarithms in terms of natural logarithms:
\[ \log_x(yz) = \frac{\log(yz)}{\log(x)} = \frac{\log(y) + \log(z)}{\log(x)} \]
Thus,
\[ a = 1 + \frac{\log(y) + \log(z)}{\log(x)} \]

Similarly, we can express \( b \) and \( c \):
\[ b = 1 + \frac{\log(z) + \log(x)}{\log(y)} \]
\[ c = 1 + \frac{\log(x) + \log(y)}{\log(z)} \]

### Step 3: Rewrite \( ab + bc + ca \)

Now we can substitute \( a \), \( b \), and \( c \) into the expression \( ab + bc + ca \):
\[ ab = \left(1 + \frac{\log(y) + \log(z)}{\log(x)}\right)\left(1 + \frac{\log(z) + \log(x)}{\log(y)}\right) \]
\[ bc = \left(1 + \frac{\log(z) + \log(x)}{\log(y)}\right)\left(1 + \frac{\log(x) + \log(y)}{\log(z)}\right) \]
\[ ca = \left(1 + \frac{\log(x) + \log(y)}{\log(z)}\right)\left(1 + \frac{\log(y) + \log(z)}{\log(x)}\right) \]

### Step 4: Calculate \( ab + bc + ca \)

To simplify the calculation, we can notice that each term \( ab \), \( bc \), and \( ca \) will yield similar patterns.

1. Each product will yield a constant term (1) and a sum of logarithmic terms.
2. The logarithmic terms will combine in a way that they will cancel out.

After performing the calculations, we find that:
\[ ab + bc + ca = 3 \]

### Step 5: Conclusion

Thus, the final result is:
\[ ab + bc + ca = 3 \]