Cho các số dương a, b khác 1 sao cho ({log _{16}}sqrt[3]{a} = {log _{{a^2}}}sqrt[9]{b} = {log _b}2)

Đặt ${\log _{16}}\sqrt[3]{a} = {\log _{{a^2}}}\sqrt[9]{b} = {\log _b}2 = t \Rightarrow \left\{ \begin{array}{l}a = {16^{3t}}\\b = {\left( 2 \right)^{\frac{1}{t}}}\\{a^{18t}} = b\end{array} \right.$

$\Rightarrow {a^{18t}} = b \Rightarrow {\left( {{{16}^{3t}}} \right)^{18t}} = {\left( 2 \right)^{\frac{1}{t}}} \Leftrightarrow {\left( 2 \right)^{216{t^2}}} = {\left( 2 \right)^{\frac{1}{t}}}$

$\Rightarrow 216{t^2} = \frac{1}{t} \Rightarrow {t^3} = \frac{1}{{216}} \Rightarrow t = \frac{1}{6} \Rightarrow \left\{ \begin{array}{l}a = {16^{\frac{1}{2}}} = 4\\b = {\left( 2 \right)^6} = 64\end{array} \right. \Rightarrow \frac{b}{a} = 16$