`sum_(n=0)^(oo) ((log_ex)^n)/(n!)`

Options

A. `log_(e) x `
B. x
C. `1/(log_(e)x)`
D. `1/x`

Answer

To solve the problem \( \sum_{n=0}^{\infty} \frac{(\log_e x)^n}{n!} \), we can recognize that this series resembles the Taylor series expansion for the exponential function \( e^x \).

### Step-by-Step Solution:

1. **Identify the Series**:
The given series is \( \sum_{n=0}^{\infty} \frac{(\log_e x)^n}{n!} \).

2. **Recall the Exponential Series**:
The Taylor series expansion for \( e^x \) is given by:
\[ e^x = \sum_{n=0}^{\infty} \frac{x^n}{n!} \]
By comparing, we can see that if we let \( x = \log_e x \), then our series becomes:
\[ e^{\log_e x} = \sum_{n=0}^{\infty} \frac{(\log_e x)^n}{n!} \]

3. **Apply the Exponential Function**:
From the properties of logarithms and exponentials, we know that:
\[ e^{\log_e x} = x \]

4. **Conclusion**:
Therefore, we conclude that:
\[ \sum_{n=0}^{\infty} \frac{(\log_e x)^n}{n!} = x \]

### Final Answer:
Thus, the final result is:
\[ \sum_{n=0}^{\infty} \frac{(\log_e x)^n}{n!} = x \]