Question

Match the inequalities with their solution sets

• A. 11,∞
• B. 1,109
• C. 1,10
• D. (199,∞)

Answer 1

The given inequalities are:
a) ${\log }_3(x-2)>2$
Now, for the value of the log to exist, $x-2>0$, or $x>2$.
Also, the base of the log is 3, which is greater than 1.
Thus, ${\log }_3(x-2)>2\Rightarrow x-2>3^2$
$\Rightarrow x-2>9$
$\Rightarrow x>11$
Thus, the solution for this inequality is $x\in (11,\infty )$

b) ${\log }_{\frac{1}{3}}(x-1)>2$
Now, for the value of the log to exist, $x-1>0$, or $x>1$.
Also, the base of the log is $\frac{1}{3}$, which is lesser than 1.
Thus, ${\log }_{\frac{1}{3}}(x-1)>2\Rightarrow x-1<{\left(\frac{1}{3}\right)}^2$
$\Rightarrow x-1<\frac{1}{9}$
$\Rightarrow x<\frac{10}{9}$
Thus, the solution for this inequality is $x\in (1,\frac{10}{9})$

c) ${\log }_3(x-1)<2$
Now, for the value of the log to exist, $x-1>0$, or $x>1$.
Also, the base of the log is 3, which is greater than 1.
Thus, ${\log }_3(x-1)<2\Rightarrow x-1<3^2$
$\Rightarrow x-1<9$
$\Rightarrow x<10$
Thus, the solution for this inequality is $x\in (1,10)$

d) ${\log }_{\frac{1}{3}}(x-2)<2$
Now, for the value of the log to exist, $x-2>0$, or $x>2$.
Also, the base of the log is $\frac{1}{3}$, which is lesser than 1.
Thus, ${\log }_{\frac{1}{3}}(x-2)<2\Rightarrow x-2>{\left(\frac{1}{3}\right)}^2$
$\Rightarrow x-2>\frac{1}{9}$
$\Rightarrow x>\frac{19}{9}$
Thus, the solution for this inequality is $x\in (\frac{19}{9},\infty )$