[tex]a = \log_{5}8\\ \\ \lg125 =\dfrac{\log_{5}125}{\log_{5}10}= \dfrac{3}{\log_{5}(5\cdot 2)}=\dfrac{3}{\log_{5}5+\log_{5}2}=\\ \\ =\dfrac{3}{1+\frac{1}{3}\log_{5}2^3}=\dfrac{3}{1+\frac{1}{3}\log_{5}8}=\dfrac{3}{1+\frac{1}{3}\cdot a}=\boxed{\dfrac{9}{3+a}}[/tex]
[tex]\\[/tex]
[tex]b=\log_{3}16\\ \\\log_{6}27 =\dfrac{\log_{3}27}{\log_{3}6} =\dfrac{3}{\log_{3}(3\cdot 2)} = \dfrac{3}{\log_{3}3+\log_{3}2} = \\ \\ = \dfrac{3}{1+\frac{1}{4}\log_{3}2^4}=\dfrac{3}{1+\frac{1}{4}\log_{3}16} = \dfrac{3}{1+\frac{1}{4}\cdot b} =\boxed{\dfrac{12}{4+b}}[/tex]
