Nếu \({2^a} = 9\) thì ( 1/16 ) ^ a /8 có

Ta có \({2^a} = 9 \Rightarrow a = {\log _2}9\) Nên \[{\left( {\frac{1}{{16}}} \right)^{\frac{a}{8}}} = {\left( {{2^{ - 4}}} \right)^{\frac{a}{8}}} = {2^{\frac{{ - a}}{2}}} = {2^{\frac{{ - {{\log }_2}9}}{2}}} = {\left( {{2^{{{\log }_2}9}}} \right)^{\frac{{ - 1}}{2}}} = {9^{\frac{{ - 1}}{2}}} = \frac{1}{3}\]