Use the approximation of $\,f(x)=\ln(1+x)\,$ at $\,x_0=0\,$ to estimate $\,\ln(15)$ I've been trying to solve an exercise from my textbook which goes as following.
For the function $f(x) = \ln(1+x)$
a) compute the Taylor's approximation at $x_0 = 0$
b) use this approximation to estimate $\ln 15$
The confusing part of the exercise for me is the point b) 
How do I exactly use the approximation at $x_0 = 0$ to approximate $\ln (15)$?
$$f(x)= \ln(1+x)$$
$$f(x)\approx x-\frac{1}{2}x^2+\frac{1}{3}x^3-\frac{1}{4}x^4$$ This is the approximation at $x_0=0$  To my understanding to approximate $\ln 15$ we need to choose a point which is close to $x=15$ as our $x_0$ in order to approximate the value of $\ln15$. So my question is how do I exactly use this formula of $f(x)=\ln(1+x)$ at $x_0=0$ for   $\ln(15)$?

The Taylor polynomial $T(x)$ that you’re using to approximate $f(x)$ around zero can not be used to approximate $f$ around $14$. Only $x$ values close to zero would work.
Please check the picture showing the two functions around zero: in red is $f(x)$ and in blue is $T(x)$. It was obtained from here

$$\ln \left ( \frac{15}{e^3}\right )=\underbrace{\ln 15-3}_{\approx -0.2919}=-\ln \left ( \frac{e^3}{15}\right )=-\ln \left ( 1+\frac{e^3-15}{15}\right )\approx -T \left ( \frac{e^3-15}{15}\right )\approx -0.2912$$
