`log5-(log25)/(2^(2))+(log125)/(3^(2))-(log625)/(4^(2))+ ……`

Options

A. log 2
B. log5
C. log`2**(log5)`
D. log 10

Answer

To solve the expression \( \log 5 - \frac{\log 25}{2^2} + \frac{\log 125}{3^2} - \frac{\log 625}{4^2} + \ldots \), we can follow these steps:

### Step 1: Rewrite the logarithmic terms
We can express the logarithmic terms in a simpler form:
- \( \log 25 = \log(5^2) = 2 \log 5 \)
- \( \log 125 = \log(5^3) = 3 \log 5 \)
- \( \log 625 = \log(5^4) = 4 \log 5 \)

Thus, we can rewrite the expression as:
\[ \log 5 - \frac{2 \log 5}{2^2} + \frac{3 \log 5}{3^2} - \frac{4 \log 5}{4^2} + \ldots \]

### Step 2: Factor out \( \log 5 \)
Now, we can factor out \( \log 5 \) from the entire expression:
\[ \log 5 \left( 1 - \frac{2}{4} + \frac{3}{9} - \frac{4}{16} + \ldots \right) \]
This simplifies to:
\[ \log 5 \left( 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \right) \]

### Step 3: Identify the series
The series inside the parentheses is:
\[ 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]
This is the alternating series for \( \log(1 + x) \) evaluated at \( x = 1 \):
\[ \log(1 + 1) = \log 2 \]

### Step 4: Substitute back into the expression
Now we can substitute back:
\[ \log 5 \cdot \log 2 \]

### Final Answer
Thus, the value of the original expression is:
\[ \log 2 \cdot \log 5 \]