Question:
If \(x > 0\), then
\[
1 + \frac{\log x}{1!} + \frac{(\log x)^2}{2!} + \cdots =
\]
If \(x > 0\), then
\[ 1 + \frac{\log x}{1!} + \frac{(\log x)^2}{2!} + \cdots = \]
\[ 1 + \frac{\log x}{1!} + \frac{(\log x)^2}{2!} + \cdots = \]
Show Hint
Recognise standard exponential series quickly.
Updated On: Mar 23, 2026
- \(x\)
- \(x^2\)
- \(2x\)
- √(x)
Show Solution
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The Correct Option is A
Solution and Explanation
Step 1: Series represents elog x.
Step 2: elog x=x
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