According to question
$$
\begin{aligned}
f(x)=f(a)+&(x-a) f^{\prime}(a) \\
+& \frac{(x-a)^{2}}{\lfloor 2} f^{\prime \prime}(a) \ldots
\end{aligned}
$$
Here \(f(x)=\log x\) and \(a=1\)
so \(f(x)=\log x \Rightarrow f(1)=0\)
\(f^{\prime}(x)=\frac{1}{x} \Rightarrow f^{\prime}(1)=1\) \(f^{\prime \prime}(x)=\frac{-1}{x^{2}} \Rightarrow f^{\prime \prime}(1)=-1\) \(f^{\prime \prime \prime}(x)=-\left(\frac{-2}{x^{3}}\right)=\frac{2}{x^{3}} \Rightarrow f^{\prime \prime \prime}(1)=2\)
So \(\log x=0+(x-1)(1)+\frac{(x-1)^{2}}{\lfloor 2}(-1)\)
$$
\begin{gathered}
+\frac{(x-1)^{3}}{\lfloor 3}(2)-\ldots \\
=(x-1)-\frac{(x-1)^{2}}{2}+\frac{(x-1)^{3}}{3}-\ldots
\end{gathered}
$$
