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$\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2 \\ & +\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots(a>0, x \in R) \text { is equal to }\end{aligned}$
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Verified Answer
The correct answer is:
$a^x$
We have,
$\begin{aligned}
& 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \\
& =1+\log _e a^x+\frac{1}{2 !}\left(\log _e a^x\right)^2+\frac{1}{3 !}\left(\log _e a^x\right)^3+\ldots \\
& =e^{\log _e a^x}=a^x
\end{aligned}$
$\begin{aligned}
& 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \\
& =1+\log _e a^x+\frac{1}{2 !}\left(\log _e a^x\right)^2+\frac{1}{3 !}\left(\log _e a^x\right)^3+\ldots \\
& =e^{\log _e a^x}=a^x
\end{aligned}$
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