The value of `log_(3)2log_(4)3log_(5)4.....log_(15)14log_(16)15` is

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. `(1)/(4)`
D. `1`

Answer

To solve the expression \( \log_{3}2 \cdot \log_{4}3 \cdot \log_{5}4 \cdots \log_{15}14 \cdot \log_{16}15 \), we can use the change of base formula for logarithms. The change of base formula states that:

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

Using this property, we can rewrite each logarithm in the product:

1. **Rewrite each logarithm:**
\[ \log_{3}2 = \frac{\log_{e}2}{\log_{e}3} \]
\[ \log_{4}3 = \frac{\log_{e}3}{\log_{e}4} \]
\[ \log_{5}4 = \frac{\log_{e}4}{\log_{e}5} \]
\[ \cdots \]
\[ \log_{16}15 = \frac{\log_{e}15}{\log_{e}16} \]

2. **Combine the logarithms:**
The product becomes:
\[ \log_{3}2 \cdot \log_{4}3 \cdot \log_{5}4 \cdots \log_{16}15 = \frac{\log_{e}2}{\log_{e}3} \cdot \frac{\log_{e}3}{\log_{e}4} \cdot \frac{\log_{e}4}{\log_{e}5} \cdots \frac{\log_{e}15}{\log_{e}16} \]

3. **Cancel terms:**
Notice that in the product, all intermediate terms will cancel out:
\[ = \frac{\log_{e}2}{\log_{e}16} \]

4. **Simplify \(\log_{e}16\):**
Since \(16 = 2^4\), we can express \(\log_{e}16\) as:
\[ \log_{e}16 = \log_{e}(2^4) = 4 \cdot \log_{e}2 \]

5. **Final expression:**
Substitute back into the equation:
\[ = \frac{\log_{e}2}{4 \cdot \log_{e}2} = \frac{1}{4} \]

Thus, the value of the original expression \( \log_{3}2 \cdot \log_{4}3 \cdot \log_{5}4 \cdots \log_{16}15 \) is:

\[ \boxed{\frac{1}{4}} \]