jmc

Writing $6=2\cdot 3$ and applying the logarithmic identity ${\log }_a(xy)={\log }_a(x)+{\log }_a(y)$ for all positive real numbers $x$, $y$, and $a$ (with $a\ne 1$), we obtain $$\begin{array}{rl}{\log }_{3}6={\log }_3(2\cdot 3)& ={\log }_{3}2+{\log }_{3}3\\ & ={\log }_{3}2+1.\end{array}$$Now use the change of base rule for logarithms on the right hand side: $$\begin{array}{rl}{\log }_{3}6& =\frac{{\log }_{4}2}{{\log }_{4}3}+1\\ & =\frac{1}{2{\log }_{4}3}+1\\ & =\frac{1}{{\log }_{4}9}+1.\end{array}$$Use the change of base rule on the right hand side once more, we get $${\log }_{3}6={\log }_{9}4+1,$$which implies that $${\log }_{4}9=1.63-1=\boxed{0.63}.$$