The value of `log_(b)a.log_(c)b.log_(a)c` is

Options

A. `0`
B. log abc
C. `1`
D. `10`

Answer

To solve the expression \( \log_b a \cdot \log_c b \cdot \log_a c \), we can use the change of base formula for logarithms. The change of base formula states that:

\[ \log_a b = \frac{\log_k b}{\log_k a} \]

for any positive base \( k \).

### Step-by-step Solution:

1. **Apply the Change of Base Formula**:
We can rewrite each logarithm using the change of base formula. Let's choose base \( e \) (natural logarithm) for simplicity.

\[ \log_b a = \frac{\log_e a}{\log_e b}, \quad \log_c b = \frac{\log_e b}{\log_e c}, \quad \log_a c = \frac{\log_e c}{\log_e a} \]

2. **Substitute Back into the Expression**:
Substitute these expressions back into the original expression:

\[ N = \log_b a \cdot \log_c b \cdot \log_a c = \left(\frac{\log_e a}{\log_e b}\right) \cdot \left(\frac{\log_e b}{\log_e c}\right) \cdot \left(\frac{\log_e c}{\log_e a}\right) \]

3. **Simplify the Expression**:
Now, we can simplify the expression by canceling out the terms:

\[ N = \frac{\log_e a}{\log_e b} \cdot \frac{\log_e b}{\log_e c} \cdot \frac{\log_e c}{\log_e a} \]

Notice that \( \log_e b \) in the numerator of the first term cancels with \( \log_e b \) in the denominator of the second term, and similarly for \( \log_e c \) and \( \log_e a \):

\[ N = \frac{\log_e a}{\log_e a} \cdot \frac{\log_e b}{\log_e b} \cdot \frac{\log_e c}{\log_e c} = 1 \]

4. **Final Result**:
Therefore, the value of \( \log_b a \cdot \log_c b \cdot \log_a c \) is:

\[ \boxed{1} \]