Solution
= Solution
Near $x=2$, use $f(x)=2\ln x$, so $f'(2)=1$ and $f''(2)=-1/2$. The <Taylor series> therefore begins
$$
\boxed{\ln(x^2)=\ln4+(x-2)-\frac14(x-2)^2+O((x-2)^3)}.
$$
The constant term is one of the requested three terms. The local expression $2\ln x$ is legitimate because this neighbourhood has $x>0$.