`1+(log_(e)5)/(1!) +(log_e5)^2/(2!)+(log_e5)^3/(3!)+ .......oo `=

Options

A. `e^(1/(log5))`
B. 5
C. `1/5`
D. `1/7`

Answer

To solve the given series

\[ 1 + \frac{\log_e 5}{1!} + \frac{(\log_e 5)^2}{2!} + \frac{(\log_e 5)^3}{3!} + \ldots \]

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 \( x \)

In our series, we can see that the variable \( x \) corresponds to \( \log_e 5 \). Therefore, we can rewrite the series as:

\[ e^{\log_e 5} \]

### Step 2: Simplify \( e^{\log_e 5} \)

Using the property of logarithms and exponentials, we know that:

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

for any positive \( a \). Hence, we can simplify:

\[ e^{\log_e 5} = 5 \]

### Conclusion

Thus, the value of the given series is:

\[ \boxed{5} \]

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