Find the value of `log_(5)10xxlog_(10)15xxlog_(15)20xxlog_(20)25`.

Options

A. A)`5//2`
B. B)5
C. C)2
D. D)`log(5/2)`

Answer

To solve the problem \( \log_{5}10 \times \log_{10}15 \times \log_{15}20 \times \log_{20}25 \), we can use the change of base formula for logarithms. The change of base formula states that:

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

for any base \( k \). We will convert all logarithms to the natural logarithm (base \( e \)).

### Step-by-Step Solution:

1. **Convert each logarithm using the change of base formula:**
\[ \log_{5}10 = \frac{\ln 10}{\ln 5} \]
\[ \log_{10}15 = \frac{\ln 15}{\ln 10} \]
\[ \log_{15}20 = \frac{\ln 20}{\ln 15} \]
\[ \log_{20}25 = \frac{\ln 25}{\ln 20} \]

2. **Substitute these into the original expression:**
\[ \log_{5}10 \times \log_{10}15 \times \log_{15}20 \times \log_{20}25 = \left(\frac{\ln 10}{\ln 5}\right) \times \left(\frac{\ln 15}{\ln 10}\right) \times \left(\frac{\ln 20}{\ln 15}\right) \times \left(\frac{\ln 25}{\ln 20}\right) \]

3. **Simplify the expression:**
Notice that \( \ln 10 \) in the numerator of the first term cancels with \( \ln 10 \) in the denominator of the second term, \( \ln 15 \) cancels with itself, and \( \ln 20 \) cancels with itself:
\[ = \frac{\ln 25}{\ln 5} \]

4. **Rewrite \( \ln 25 \):**
We know that \( 25 = 5^2 \), so:
\[ \ln 25 = \ln(5^2) = 2 \ln 5 \]

5. **Substitute this back into the expression:**
\[ = \frac{2 \ln 5}{\ln 5} \]

6. **Cancel \( \ln 5 \):**
\[ = 2 \]

### Final Answer:
The value of \( \log_{5}10 \times \log_{10}15 \times \log_{15}20 \times \log_{20}25 \) is \( 2 \).