The coefficient of x^3 in the expansion of 3^ | vedclass

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(D) We know that $3^x = e^{\log(3^x)} = e^{x \log 3}$.Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$,wh

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$\frac{(\log 3)^3}{6}$

Vedclass · Answer

(D) We know that $3^x = e^{\log(3^x)} = e^{x \log 3}$.Using the Taylor series expansion for $e^u = 1 + u + \frac{u^2}{2!} + \frac{u^3}{3!} + \dots$,where $u = x \log 3$:$3^x = 1 + \frac{x \log 3}{1!} + \frac{(x \log 3)^2}{2!} + \frac{(x \log 3)^3}{3!} + \dots$$3^x = 1 + x \log 3 + \frac{x^2 (\log 3)^2}{2} + \frac{x^3 (\log 3)^3}{6} + \dots$The coefficient of $x^3$ is $\frac{(\log 3)^3}{6}$.

The coefficient of $x^3$ in the expansion of $3^x$ is

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