= Solution
Write $q(x)=a+bx$. <Orthogonality> of $x^2-q$ to $1,x$ gives
$$
\begin{pmatrix}\log3&2\\2&4\end{pmatrix}\binom ab=\binom4{26/3}.
$$
The entries are obtained by integrating $1/x,1,x,x^2$ over the given <interval>. Subtracting the second equation divided by two from the first gives $(\log3-1)a=-1/3$. Thus
$$
\boxed{q(x)=-\frac1{3(\log3-1)}+\left(\frac{13}6+\frac1{6(\log3-1)}\right)x.}
$$
The <Gram matrix> is positive definite, with determinant $4(\log3-1)>0$, so this <orthogonal projection> is the unique <best approximation>.
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