Question

$x^{lo{g}_{9x}}>9$ 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 +ive.
$\therefore lo{g}_{9}x<-1or>1$
$\therefore x<9^{-1}$ or $>9^1$ or $x<\frac{1}{9}or>9$
Since x + ive $\therefore x\epsilon (0,\frac{1}{9})\cup (9,\infty )$