Question

Solve for $x$:

$3^{{\log }_3\sqrt{x-1}}<3^{{\log }_3(x-6)}+3$

• A. $(5,+\infty )$
• B. $(6,+\infty )$
• C. $(1,2)\cup (3,5)$
• D. $(-\infty ,2)\cup (5,\infty )$

Answer 1

Answer: (B)

$(6,+\infty )$

$3^{{\log }_3\sqrt{x-1}}<3^{{\log }_3(x-6)}+3$

$\Rightarrow {\sqrt{x-1}}^{{\log }_{3}3}<(x-6{)}^{{\log }_{3}3}+3$

$\Rightarrow \sqrt{x-1}<(x-6)+3$

$\Rightarrow \sqrt{x-1}<x-3$

Squaring both sides,

$(x-1)<(x-3{)}^2$

$\Rightarrow x-1<x^2-6x+9$

$\Rightarrow x^2-7x+10>0\Rightarrow (x-5)(x-2)>0$

$\Rightarrow xϵ(-\infty ,2)\cup (5,\infty )$

But, $x-6>0\Rightarrow x>6$

$\therefore xϵ(6,\infty )$