By taylor series,

$f(x)= f(a)+ (x-a)f'(a)+ \dfrac {(x-a)^2}{2!}f'' (a)+ \dfrac {(x-a)^3}{3!} f'''(a)+ \ \cdots \ \cdots $

Here, a=0

$f(x)= f(0)+xf' (0)+ \dfrac {x^2}{2!}f''(a)+ \dfrac {x^3}{3!}f'''(a)+ \ \cdots \ \cdots (A)$

$f(x)=\log (1+x)$

$ f(0)= \log (1+0)=0 $

$ f'(x)= \dfrac {1}{1+x} $

$f'(0)= \dfrac {1}{1+0}= 1 $

$f''(x) = - \dfrac {1}{(1+x)^2} \\ f''(0) = -\dfrac {1}{(1+0)^2}= -1 \\ f'''(x)= \dfrac {2}{(1+x)^3} \\ f'''(0) = \dfrac {2}{(1+0)^3}=2 $

Plugging all the values in (A),

$f(x)=0 +x(1)+ \dfrac {x^2}{2}(-1)+ \dfrac {x^3}{6}(2)- \cdots \ \cdots $

$\log (1+x)= x- \dfrac {x^2}{2}+ \dfrac {x^3}{3}- \cdots \ \cdots $

To get log x plug in $x=x-1,$

$\log (x)= (x-1)- \dfrac {(x-1)^2}{2}+ \dfrac {(x-1)^3}{3}- \ \cdots \ \cdots $