Substitution in the taylor series of $\ln(1+ x)$

Question

We know that $$\ln(1+ x) = \sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n}$$ I need to find the Maclaurin series for $\ln(4 - x)$. The approach I have tried is writing $\ln(4 - x)$ as $\ln(1 + (3 - x))$ and then substituting $3-x$ in $$\sum_{n=1}^\infty\frac{(-1)^{n+1}x^n}{n} \rightarrow \sum_{n=1}^\infty\frac{(-1)^{n+1}(3-x)^n}{n}$$

Is this a correct approach, since the answer in the textbook I am referring to uses derivatives to find the series.

Answer 1

Hint

$$\log(4-x)=\log(4)+\log \left(1-\frac{x}{4}\right)$$ So, use you expression for $\log(1-t)$ and replace $t$ by $\frac x4$.

Answer 2

You are also right but your series will be valid only for $$|3-x|<1 \implies -1<3-x <1 \implies 2<x< 4.$$ Your's is a Taylor series about $x=3$, which can also be found as usual by finding various derivatives of $\log(4-x)$ at $x=3$. Series suggested by @Claude Lebovici is the actual Maclaurin series which is also a Taylor-series about $x=0$.