`1+ 2 (log_ea)+(2^2)/(2!) (log_ea)^2 +(2^3)/(3!) (log_e a)^3+...=`

Options

A. e
B. a
C. `a^2`
D. `e^2`

Answer

To solve the given series:

\[ 1 + 2 (\log_e a) + \frac{2^2}{2!} (\log_e a)^2 + \frac{2^3}{3!} (\log_e a)^3 + \ldots \]

we can recognize that this series resembles the Taylor series expansion for the exponential function \( e^x \), which is given by:

\[ e^x = 1 + \frac{x}{1!} + \frac{x^2}{2!} + \frac{x^3}{3!} + \ldots \]

### Step 1: Identify the variable in the series
In our series, we can see that the term \( 2 (\log_e a) \) can be treated as \( x \) in the exponential series. Thus, we can rewrite the series as:

\[ 1 + \frac{(2 \log_e a)}{1!} + \frac{(2 \log_e a)^2}{2!} + \frac{(2 \log_e a)^3}{3!} + \ldots \]

### Step 2: Recognize the series as an exponential function
This series can be recognized as the expansion of \( e^{x} \) where \( x = 2 \log_e a \):

\[ e^{2 \log_e a} = 1 + \frac{(2 \log_e a)}{1!} + \frac{(2 \log_e a)^2}{2!} + \frac{(2 \log_e a)^3}{3!} + \ldots \]

### Step 3: Simplify the exponential expression
Using the property of logarithms, we know that:

\[ e^{\log_e b} = b \]

Thus, we can simplify \( e^{2 \log_e a} \):

\[ e^{2 \log_e a} = a^2 \]

### Conclusion
Therefore, the value of the series is:

\[ 1 + 2 (\log_e a) + \frac{2^2}{2!} (\log_e a)^2 + \frac{2^3}{3!} (\log_e a)^3 + \ldots = a^2 \]

### Final Answer
The answer is \( a^2 \).

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