Question

$x^{lo{g}_{9}x}>9$ implies

• A. $x\epsilon (0,\infty )$
• B. $x\epsilon (0,\frac{1}{9})\cup (9,\infty )$
• C. $x>1$
• D. $x<2$

Answer 1

The correct option is B

$x\epsilon (0,\frac{1}{9})\cup (9,\infty )$

Change from exponential to log.
$lo{g}_{9}x>lo{g}_{x}9=\frac{1}{lo{g}_{9}x}$
or $(lo{g}_{9}x{)}^2-1>0$
or $(y+1)(y-1)>0$ where $y=lo{g}_{9}x$
Where $x$ is positive.
$\therefore lo{g}_{9}x<-1or>1$
$\therefore x<9^{-1}$ or $>9^1$ or $x<\frac{1}{9}or>9$
So $x$ is positive in $x\epsilon (0,\frac{1}{9})\cup (9,\infty )$