Find Maclaurin expansion of $y=2^x$ to $x^4$

Question

Find Maclaurin expansion of $$y=2^x\text{ to } x^4$$

This is my try.

We have $\displaystyle 2^x=e^{x\ln 2} =\left[1+\frac{x^2}2+\frac{x^3}6+\frac{x^4}{24}+o(x^4)\right]^{\ln 2}$ with $o(x^4)$ is infinitesimal larger, but I think that it doesn't meet the problem because $\left(x^4\right)^{\ln 2}$ is smaller than $x^4$.

Answer 1

It is much simpler: by substitution, we get $$2^x=e^{x\ln 2} =1+x\ln 2+\frac{x^2\ln^22}2+\frac{x^3\ln^32}6+\frac{x^4\ln^42}{24}+o(x^4).$$

Answer 2

You didn't apply the taylor series properly.

We have $$\displaystyle 2^x=e^{(x\ln 2)} =1+\frac{(x\ln2)}{1!}+\frac{(x\ln2)^2}{2!}+\frac{(x\ln2)^3}{3!}+\frac{(x\ln2)^4}{4!}+O(x^5)$$