`e^{(x-1)-1/2(x-1)^2+((x-1)^3)/3-(x-1)^(4)/4+......}` is eqaul to

Options

A. `log(x-1)`
B. `logx`
C. x
D. none of these

Answer

To solve the problem, we need to analyze the expression given:

\[ e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \ldots} \]

This expression can be recognized as the Taylor series expansion for the natural logarithm function. The series expansion for \(\log(1 + x)\) is given by:

\[ \log(1 + x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \ldots \]

### Step 1: Identify the series
In our case, we can see that the series matches the form of \(\log(1 + (x - 1))\) where \(x - 1\) is substituted for \(x\). Thus, we can rewrite the series as:

\[ \log(1 + (x - 1)) = \log(x) \]

### Step 2: Substitute into the exponential function
Now we have:

\[ e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \ldots} = e^{\log(x)} \]

### Step 3: Simplify using properties of logarithms
Using the property that \(e^{\log(a)} = a\), we can simplify this to:

\[ e^{\log(x)} = x \]

### Conclusion
Thus, the expression simplifies to:

\[ e^{(x-1) - \frac{1}{2}(x-1)^2 + \frac{(x-1)^3}{3} - \frac{(x-1)^4}{4} + \ldots} = x \]

### Final Answer
The final answer is:

\[ \boxed{x} \]