Find the terms of the polynomial. Manually, if you're not familiar with the power series for \(\ln x\).
\[\begin{matrix}
f(x)=\ln x&&&f(0.845)\approx-0.1684\\
f'(x)=\frac{1}{x}&&&f'(0.845)\approx1.1834\\
f''(x)=-\frac{1}{x^2}&&&f''(0.845)\approx -1.4005\\
f^{(3)}(x)=\frac{2}{x^3}&&&f^{(3)}(0.845)\approx3.3148\\
f^{(4)}(x)=-\frac{6}{x^4}&&&f^{(4)}(0.845)\approx -11.7686
\end{matrix}\]
The Taylor polynomial of order 4 will give you an approximation, i.e.
\[\ln x\approx \sum_{i=0}^4\frac{f^{(n)}(0.845)}{n!}(x-0.845)^n\]
Here's where I start to kick myself for doing more work than necessary. Notice that the \(i=1,2,3,4\) terms disappear because when you plug in \(x=0.845\), the terms disappear, leaving you with the \(i=0\) term. So you have
\[\ln 0.845\approx -0.1684\]