TS EAMCET · Maths · Basic of Mathematics
\(\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2 \ & +\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots(a>0, x \in R) \text { is equal to }\end{aligned}\)
- A \(a\)
- B \(a^x\)
- C \(a^{\log _e x}\)
- D \(x\)
Answer & Solution
Correct Answer
(B) \(a^x\)
Step-by-step Solution
Detailed explanation
We have, \(\begin{aligned} & 1+x \log _e a+\frac{x^2}{2 !}\left(\log _e a\right)^2+\frac{x^3}{3 !}\left(\log _e a\right)^3+\ldots \\ & =1+\log _e a^x+\frac{1}{2 !}\left(\log _e a^x\right)^2+\frac{1}{3 !}\left(\log _e a^x\right)^3+\ldots \\ & =e^{\log _e a^x}=a^x \end{aligned}\)
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