Question

(i) Solving $1+lo{g}_2(x-1)\le lo{g}_{x-1}4$ we get domain of x be $(m,n]$.Find $m+n$?

Answer 1

$\Rightarrow 1+{\log }_2(x-1)\le {\log }_{(x-1)}4\Rightarrow {\log }_{2}2+{\log }_2(x-1)\le {\log }_{(x-1)}4\Rightarrow {\log }_{2}2(x-1)-{\log }_{(x-1)}4\le 0$
So substituting
$x-1=2\Rightarrow x=3{\log }_{2}4-{\log }_{2}4=0$
So required values of $x\in (1,3]$
$\Rightarrow m+n=4$