Question

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

Answer 1

$a={\log }_3{\log }_{3}2\Rightarrow 3^a={\log }_{3}2\Rightarrow 3^{-a}={\log }_{2}3$
Now,
$2^{(3^{-a}-k)}=2^{{\log }_{2}3}\cdot 2^{-k}=3\cdot 2^{-k}$

$\because 1<2^{(3^{-a}-k)}<2\therefore 1<3\cdot 2^{-k}<2\Rightarrow \frac{1}{3}<2^{-k}<\frac{2}{3}$
$\Rightarrow 3>2^k>\frac{3}{2}$

Clearly, this inequality will hold good only for $k=1$.