Question

If the least integral value satisfying the equation ${\log }_3\sqrt{x^2-4x+4}=2^{{\log }_2\left({\log }_3(|x|-2)\right)}$ is $\alpha$, then find the number of zeroes after decimal and before first significant digit in the number of $(\alpha {)}^{-4\alpha }$.

Answer 1

Answer: 9

${\log }_3|x-2|=2^{{\log }_2\left({\log }_3(|x|-2)\right)}$
$\Rightarrow {\log }_3|x-2|={\log }_3(|x|-2)$
$\Rightarrow |x-2|=|x|-2$
$\Rightarrow |x|-|x-2|=2$
$\text{ Case-I }:x<0$
$-x+x-2=0$
$-2=2\text{ (no solution) }$
Case-II: $0\le x<2$
$x+(x-2)=2$
$\Rightarrow 2x=4\Rightarrow x=2\text{ (no solution) }$
$\text{ Case-III }:x\ge 2$
$x-(x-2)=2$
$\Rightarrow 2=2$
$\therefore x\ge 2$
But ${\log }_3(|x|-2)>0\Rightarrow |x|-2>1\Rightarrow |x|>3$
$\therefore x\in (3,\infty )$ is the solution set of the equation
Least integral value of $x$ is 4 .
Now,$(\alpha {)}^{-4\alpha }=(4{)}^{-16}=2^{-32}$
$N=2^{-32}$
${\log }_{10}N=-32{\log }_{10}2=-32\times 0.301=-9.632$
$\therefore$ Number of zeros after decimal and before first significant digit in $2^{-32}$ is 9