Question

Let $a={\log }_3{\log }_{3}2.$ An integer k satisfying $1<2^{\left(-k+3^{-a}\right)}<2,$ is

Answer 1

$a={\log }_3{\log }_{3}2\Rightarrow 3^a={\log }_{3}2\Rightarrow 3^{-a}=\frac{1}{{\log }_{3}2}={\log }_{2}3\mathrm{Now},1<2^{\left(-k+3^{-a}\right)}<2\Rightarrow 1<2^{\left(-k+{\log }_{2}3\right)}<2\Rightarrow 1<2^{-k}\cdot 3<2\Rightarrow \frac{1}{3}<2^{-k}<\frac{2}{3}\Rightarrow \frac{3}{2}<2^k<3\Rightarrow {\log }_2\frac{3}{2}<k<{\log }_{2}3\mathrm{Hence},\mathrm{the}\mathrm{integral}\mathrm{value}\mathrm{of}k\mathrm{is}1.$