Question

The set of real values of x for which $2^{{log}_{\sqrt{2}}(x-1)}>x+5$ is

• A. $(-\infty ,-1)\cup (4,+\infty )$
• B. $(4,+\infty )$
• C. $(-1,4)$
• D. $Noneofthese$

Answer 1

The correct option is B $(4,+\infty )$

$2^{{log}_{\sqrt{2}}(x-1)}>x+5\Rightarrow 2^{lo{g}_2(x-1{)}^2}>x+5\Rightarrow (x-1{)}^2>x+5\Rightarrow x^2-3x-4>0\Rightarrow (x-4\left)\right(x+1)>0\Rightarrow x>4orx<-1$
But for given log to be defined, x - 1 > 0
$i.e.,x>1\therefore x>4\Rightarrow xϵ(4,\infty ).$