If ${\log }_{12} 27=\mathrm{a},$then ${\log }_6 16=$

$\mathrm{a}=\frac{\log 27}{\log 12}=\frac{3 \log 3}{\log 3+2 \log 2}\Rightarrow \log 3=\frac{2\mathrm{a} \log 2}{3-\mathrm{a}}$

${\log }_6 16=\frac{\log 16}{\log 6}=\frac{4 \log 2}{\log 2+\log 3}$

$=\frac{4 \log 2}{\log 2+{\displaystyle \frac{2a \log 2}{3-\mathrm{a}}}}=\frac{4(3-\mathrm{a})}{3-\mathrm{a}+2\mathrm{a}}=4.\frac{3-\mathrm{a}}{3+\mathrm{a}}$