Taylor series of $\ln{\sqrt[4]{\frac{x-2}{5-x}}}$ to $o((x-x_0)^n)$ when $x_0 = 3$ Well I have tried to get it as
$$f(x) = f(x_0) + \frac{f'(x_0)(x-x_0)}{1!} + \frac{f''(x_0)(x-x_0)^2}{2!} + ... + o((x-x_0)^n)$$
and got wrong results:
First: $$f'(x) = \frac{3}{4(x-2)(5-x)}$$ 
Second: $$f''(x) = \frac{3(2x-7)}{4(x-2)^2(5-x)^2} , $$
Third: $$f'''(x) = \frac{3}{4}*\frac{2(x-2)(5-x) - 2(5-x)(2x-7) + 2(x-2)(2x-7)}{(x-2)^3(5-x)^3}$$
Then I have tried changing it to $$\ln{\sqrt[4]{\frac{x-2}{5-x}}} = \frac{ln{(x-2)}}{4} - \frac{ln{(5-x)}}{4}$$ and got nothing useful. So I need an advice how to start. Or at least a hint.

Answer is $$-\frac{ln2}{4} + \sum^n_{k=1}{\frac{(-1)^{k-1} + 2^{-k}(x-3)^k}{4k}} + o((x-3)^n)$$
