`log_(a)b xx log_(b)c xx log_(c) d xx log_(d)a` is equal to

Options

A. 2
B. 3
C. 1
D. 4

Answer

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

\[ \log_x y = \frac{\log_k y}{\log_k x} \]

for any positive base \( k \).

### Step-by-Step Solution:

1. **Apply Change of Base Formula:**
We can rewrite each logarithm using the natural logarithm (or any common base). Let's use natural logarithm \( \ln \):

\[ \log_a b = \frac{\ln b}{\ln a}, \quad \log_b c = \frac{\ln c}{\ln b}, \quad \log_c d = \frac{\ln d}{\ln c}, \quad \log_d a = \frac{\ln a}{\ln d} \]

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

\[ \log_a b \times \log_b c \times \log_c d \times \log_d a = \left(\frac{\ln b}{\ln a}\right) \times \left(\frac{\ln c}{\ln b}\right) \times \left(\frac{\ln d}{\ln c}\right) \times \left(\frac{\ln a}{\ln d}\right) \]

3. **Combine the Fractions:**
When we multiply these fractions, we can combine them:

\[ = \frac{\ln b \cdot \ln c \cdot \ln d \cdot \ln a}{\ln a \cdot \ln b \cdot \ln c \cdot \ln d} \]

4. **Cancel Common Terms:**
Notice that the numerator and the denominator are the same, so they cancel out:

\[ = 1 \]

5. **Conclusion:**
Therefore, the value of the original expression is:

\[ \log_a b \times \log_b c \times \log_c d \times \log_d a = 1 \]

### Final Answer:
The final answer is \( 1 \).