Question

If ${\log }_2(x-2)<{\log }_4(x-2)$, then the interval in which $x$ lies is

• A. $xϵ(2,3)$
• B. $xϵ(2,3]$
• C. $xϵ[2,3)$
• D. $xϵ[2,3]$

Answer 1

The correct option is A $xϵ(2,3)$

Here, $(x-2)>0\Rightarrow x>2$ ...........(1)

${\log }_2(x-2)<{\log }_4(x-2)\Rightarrow {\log }_2(x-2)<{\log }_{2^2}(x-2)$

Using the law,${\log }_{b^n}{a}^m=\frac{m}{n}{\log }_{b}a$ we have

${\log }_2(x-2)<\frac{1}{2}{\log }_2(x-2)$

$\Rightarrow {\log }_2(x-2)-\frac{1}{2}{\log }_2(x-2)=d\frac{1}{2}{\log }_2(x-2)<0$

$\Rightarrow {\log }_2(x-2)<0$

Using the definition of logarithm, we have
$x-2<2^0=1$

$\therefore ,x<1+2=3$ ..........(2)

From (1) and (2), $2<x<3$ or $xϵ(2,3)$