`log_(4)2-log_(8)2+log_(16)2-....=`

Options

A. `e^(2)`
B. `log_(e )2+1`
C. `log_(e )3-2`
D. `1-log_(e) 2`

Answer

To solve the expression \( \log_{4}2 - \log_{8}2 + \log_{16}2 - \ldots \), we can follow these steps:

### Step 1: Rewrite the logarithms in terms of base 2
Using the change of base formula, we can express each logarithm in terms of base 2:
\[ \log_{4}2 = \frac{\log_{2}2}{\log_{2}4}, \quad \log_{8}2 = \frac{\log_{2}2}{\log_{2}8}, \quad \log_{16}2 = \frac{\log_{2}2}{\log_{2}16} \]
Since \( \log_{2}2 = 1 \), we have:
\[ \log_{4}2 = \frac{1}{\log_{2}4}, \quad \log_{8}2 = \frac{1}{\log_{2}8}, \quad \log_{16}2 = \frac{1}{\log_{2}16} \]

### Step 2: Substitute the values of logarithms
Now we can substitute the values:
\[ \log_{2}4 = 2, \quad \log_{2}8 = 3, \quad \log_{2}16 = 4 \]
Thus, we can rewrite the expression:
\[ \log_{4}2 = \frac{1}{2}, \quad \log_{8}2 = \frac{1}{3}, \quad \log_{16}2 = \frac{1}{4} \]
So the expression becomes:
\[ \frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \ldots \]

### Step 3: Identify the pattern
The series can be expressed as:
\[ \sum_{n=1}^{\infty} (-1)^{n+1} \frac{1}{n+1} \]
This series starts from \( n=1 \) and alternates signs.

### Step 4: Recognize the series as a logarithmic series
This series can be recognized as related to the Taylor series expansion of \( \log(1+x) \). Specifically, we can relate it to:
\[ \log(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]
If we set \( x = 1 \), we find:
\[ \log(2) = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + \ldots \]

### Step 5: Combine the series
Thus, we can rewrite our series as:
\[ \frac{1}{2} - \left( \frac{1}{3} - \frac{1}{4} + \ldots \right) \]
This can be simplified to:
\[ \frac{1}{2} - \log(2) \]

### Final Result
The final result of the original expression is:
\[ \frac{1}{2} - \log(2) \]