The value of `log_(3)4.log_(4)5.log_(5)6.log_(6)7log_(7)8.log_(8)9` is

Options

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

Answer

To solve the expression \( \log_{3}4 \cdot \log_{4}5 \cdot \log_{5}6 \cdot \log_{6}7 \cdot \log_{7}8 \cdot \log_{8}9 \), we can use the change of base formula for logarithms, which states:

\[ \log_{a}b = \frac{\log b}{\log a} \]

### Step-by-Step Solution:

1. **Rewrite each logarithm using the change of base formula**:
\[ \log_{3}4 = \frac{\log 4}{\log 3}, \quad \log_{4}5 = \frac{\log 5}{\log 4}, \quad \log_{5}6 = \frac{\log 6}{\log 5}, \quad \log_{6}7 = \frac{\log 7}{\log 6}, \quad \log_{7}8 = \frac{\log 8}{\log 7}, \quad \log_{8}9 = \frac{\log 9}{\log 8} \]

2. **Substitute these into the original expression**:
\[ \log_{3}4 \cdot \log_{4}5 \cdot \log_{5}6 \cdot \log_{6}7 \cdot \log_{7}8 \cdot \log_{8}9 = \left(\frac{\log 4}{\log 3}\right) \cdot \left(\frac{\log 5}{\log 4}\right) \cdot \left(\frac{\log 6}{\log 5}\right) \cdot \left(\frac{\log 7}{\log 6}\right) \cdot \left(\frac{\log 8}{\log 7}\right) \cdot \left(\frac{\log 9}{\log 8}\right) \]

3. **Notice that most terms will cancel**:
\[ = \frac{\log 4}{\log 3} \cdot \frac{\log 5}{\log 4} \cdot \frac{\log 6}{\log 5} \cdot \frac{\log 7}{\log 6} \cdot \frac{\log 8}{\log 7} \cdot \frac{\log 9}{\log 8} \]
Here, \( \log 4 \) in the numerator of the first term cancels with \( \log 4 \) in the denominator of the second term, and this pattern continues.

4. **After cancellation, we are left with**:
\[ = \frac{\log 9}{\log 3} \]

5. **Simplify further**:
\[ = \log_{3}9 \]

6. **Since \( 9 = 3^2 \)**:
\[ \log_{3}9 = \log_{3}(3^2) = 2 \]

### Final Answer:
The value of \( \log_{3}4 \cdot \log_{4}5 \cdot \log_{5}6 \cdot \log_{6}7 \cdot \log_{7}8 \cdot \log_{8}9 \) is \( 2 \).