# How to represent $\ln(5-x)$ as a power series?

I know that $$\ln(1+x)=\sum _{n=1}^{\infty }\:\left(-1\right)^{n-1}\frac{x^{n}}{n}$$

Hint: $$\ln(5-x)=\ln\left[5\left(1-\frac{x}{5}\right)\right]=\ln 5+\ln\left(1-\frac{x}{5}\right).$$
• But then we use $\sum _{n=0}^{\infty }\:\frac{\left(\frac{x}{5}\right)^{n-1}}{n-1}\left(-1\right)^n$ instead of $\sum _{n=0}^{\infty }\:\frac{\left(x\right)^{n-1}}{n-1}\left(-1\right)^n$ ? – DDDD Jun 11 '14 at 14:43
In your case it's more convenient to use $$\log \bigg( \frac{1}{1-w} \bigg) = \sum_{k=1}^{n} \frac{w^k}{k}$$ just multiply by $-1$ and set $w=5x$.