`sum_(n=0)^(oo) ((log_(e)X)^n)/(n!)` is equal to

Options

A. `log_(e) x`
B. `x`
C. `log_(x) e`
D. none of these

Answer

To solve the problem, we need to evaluate the infinite series:

\[ \sum_{n=0}^{\infty} \frac{(\log_e X)^n}{n!} \]

This series resembles the Taylor series expansion for the exponential function \(e^x\), which is given by:

\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]

### Step 1: Identify the variable
In our case, we can identify \(x\) with \(\log_e X\). Thus, we can rewrite the series as:

\[ \sum_{n=0}^{\infty} \frac{(\log_e X)^n}{n!} = e^{\log_e X} \]

### Step 2: Simplify using properties of logarithms
Using the property of exponents and logarithms, we know that:

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

### Final Result
Therefore, the value of the infinite series is:

\[ \sum_{n=0}^{\infty} \frac{(\log_e X)^n}{n!} = X \]

### Summary of Steps
1. Recognize the series as the Taylor series for \(e^x\).
2. Substitute \(x\) with \(\log_e X\).
3. Simplify using the property \(e^{\log_e X} = X\).