If `x gt 0,` then `1+(log_(e^2)x)/(1!)+(log_(e^2)x)^2/(2!).....`

Options

A. `x`
B. `x^(2)`
C. 2x
D. `sqrt(x)`

Answer

To solve the problem, we need to evaluate the expression:

\[ 1 + \frac{\log_{e^2} x}{1!} + \frac{(\log_{e^2} x)^2}{2!} + \frac{(\log_{e^2} x)^3}{3!} + \ldots \]

This expression can be recognized as the Taylor series expansion for the exponential function \( e^x \), where \( x \) is replaced by \( \log_{e^2} x \).

### Step-by-step Solution:

1. **Recognize the Series**: The given series can be rewritten in the form of the exponential function's Taylor series:

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

Here, we can see that \( x = \log_{e^2} x \).

2. **Substituting \( x \)**: Substitute \( x \) in the Taylor series:

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

3. **Using the Change of Base Formula**: We can express \( \log_{e^2} x \) using the change of base formula:

\[ \log_{e^2} x = \frac{\log_e x}{\log_e e^2} = \frac{\log_e x}{2} \]

4. **Substituting Back**: Substitute this back into the exponential function:

\[ e^{\log_{e^2} x} = e^{\frac{\log_e x}{2}} \]

5. **Simplifying the Exponential**: Using the property of exponents:

\[ e^{\frac{\log_e x}{2}} = x^{\frac{1}{2}} = \sqrt{x} \]

Thus, the final result is:

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

### Final Answer:
\[ \sqrt{x} \]