Question

If A is the set of all $xϵR$ such that $x^{(logx{)}^2-3logx+1}>1000$, and $A=(a,\infty )$ then $\sqrt{10a}$ will be ___. (Base of $logx$ is 10).

Answer 1

$A=(a,\infty )$
$(logx{)}^2-3logx+1>lo{g}_x{10}^3=3lo{g}_{x}10$
If $lo{g}_{10}x=t$ then we have
$t^2-3t+1>\frac{3}{t}$
or $t^3-3t^2+t-3>0$
or $t(t^2+1)-3(t^2+1)>0$
or $(t^2+1)(t-3)>0\Rightarrow t-3>0$
as $t^2+1$ is always + ive
$\therefore t>3$ or $lo{g}_{10}x>3$
$\therefore x>{10}^3=1000$

$\therefore xϵ(1000,\infty )$

or $a=1000$