Question

Determine domain $\log \{{\left(\log x\right)}^2-5\log x+6\}$.
The domain can be expressed as $\therefore xϵ(0,{10}^a)\cup ({10}^b,\infty )$, then a+b = ?

Answer 1

$\log \{{\left(\log x\right)}^2-5\log x+6\}$.
From the given inequality, it follows that
${\left(\log x\right)}^2-5\log x+6>0$
Put $\log x=t$
$\Rightarrow t^2-5t+6>0$
$\Rightarrow (t-2)(t-3)>0$
$\Rightarrow t<2$ or $t>3$
$\Rightarrow \log x<2$ or $\log x>3$
$\Rightarrow x<{10}^2$ or $x>{10}^3$
Since, $\log x$ is defined for $x>0$
$\Rightarrow x\in (0,{10}^2)\cup ({10}^3,\infty )$
Comparing with given domain ,
$a={10}^2,b={10}^3$
$\Rightarrow a+b={10}^2+{10}^3$