Question

Solve the equation for $x$: ${\log }_{{\log }_2\left(x/2\right)}(x^2-10x+22)>0$.

• A. $x\in (-\infty ,5-\sqrt{3})\cup (5+\sqrt{3},7)$
• B. $x\in (-\infty ,5-\sqrt{3})\cup (5+\sqrt{3},\infty )$
• C. $x\in (3,5-\sqrt{3})\cup (7,\infty )$
• D. $x\in (3,5-\sqrt{3})\cup (5+\sqrt{3},\infty )$

Answer 1

Answer: (C), (D)

$x\in (3,5-\sqrt{3})\cup (5+\sqrt{3},\infty )$
$x\in (3,5-\sqrt{3})\cup (7,\infty )$

$lo{g}_{lo{g}_2\left(x/2\right)}(x^2-10x+22)>0$
$\therefore \frac{log(x^2-10x+22)}{\frac{log\left(\frac{x}{2}\right)}{log2}}$ > 0
$\therefore log(x^2-10x+22)>0$
But, if $log(x^2-10x+22)>0$ then, the value of $x^2-10x+22$ will have to be >0
Now this part we will see bit later but for now,
Lets solve, $log(x^2-10x+22)>0$
Now because $log1=0$
$\therefore x^2-10x+22>1$
$\therefore x^2-10x+21>0$
$\therefore (x-7)(x-3)>0$
$\therefore$ either$x>7$ or $x>3$ i.e., Range of x is 3 to $\infty$ and 7 to $\infty$
Now, Lets solve for, $x^2-10x+22>0$
$\therefore x>5+\sqrt{3}$ or $x>5-\sqrt{3}$
So, range of x is $5-\sqrt{3}$ to $\infty$ & $5+\sqrt{3}$ to $\infty$
Hence, $x\in (3,5-\sqrt{3})\cup (7,\infty )$
And $x\in (3,5-\sqrt{3})\cup (5+\sqrt{3},\infty )$