Answer
$a-\\
\log_{8}27=\frac{\log_{2}27}{\log_{2}8}=\frac{\log_{2}3^{3}}{\log_{2}2^{3}}=\frac{3\log_{2}3}{3\log_{2}2}
=\frac{\log_{2}3}{\log_{2}2}=\frac{\log_{2}3}{1}=\log_{2}3 \\
b-\\
\log_{16}9=\frac{\log_{4}9}{\log_{4}16}=\frac{\log_{4}3^{2}}{\log_{4}4^{2}}=\frac{2\log_{4}3}{2\log_{4}4}
=\frac{\log_{4}3}{\log_{4}4}=\frac{\log_{4}3}{1}=\log_{4}3 \\
(we\,use\,\log_{y}x=\frac{\log_{a}x}{\log_{a}y}\\\log_{y}x^{n}=n\log_{y}x \\\log_{x}x=1)
$
Work Step by Step
$a-\\
\log_{8}27=\frac{\log_{2}27}{\log_{2}8}=\frac{\log_{2}3^{3}}{\log_{2}2^{3}}=\frac{3\log_{2}3}{3\log_{2}2}
=\frac{\log_{2}3}{\log_{2}2}=\frac{\log_{2}3}{1}=\log_{2}3 \\
b-\\
\log_{16}9=\frac{\log_{4}9}{\log_{4}16}=\frac{\log_{4}3^{2}}{\log_{4}4^{2}}=\frac{2\log_{4}3}{2\log_{4}4}
=\frac{\log_{4}3}{\log_{4}4}=\frac{\log_{4}3}{1}=\log_{4}3 \\
(we\,use\,\log_{y}x=\frac{\log_{a}x}{\log_{a}y}\\\log_{y}x^{n}=n\log_{y}x \\\log_{x}x=1)
$