Question

Solve the given inequality ${{\log }_{x}}0.5>2$ .
(a)$\left( \sqrt{\dfrac{1}{2}},1 \right)$
(b)$\left( -\infty ,1 \right)$
(c)$\left( -1,0 \right)$
(d)$\left( -1,1 \right)$

Answer 1

Hint: For solving the inequality given in the question, you need to make two cases, one where x>1 and the other where x<1. So, for the case where x>1, ${{\log }_{x}}0.5>2$ implies $0.5>{{x}^{2}}$ , while in case of x<1 ${{\log }_{x}}0.5>2$ implies $0.5<{{x}^{2}}$ . Finally, take the union of the values of x of the two cases to get the final answer.

Complete step-by-step answer:
The given inequality is:
${{\log }_{x}}0.5>2$
So, looking at the terms, we know that the base of a logarithmic function must be a positive real number other than 1. So, we can say that x cannot be $(-\infty ,0]\cup \\{1\\}$ .
Now we will make two cases, one where x>1 and the other where x<1. First we will solve for x>1.
So, for x>1, ${{\log }_{x}}0.5>2$ implies $0.5>{{x}^{2}}$. As in this case x is greater than 1 and we know that square of any number greater than 1 is always greater than 1. So, there are no possible values of x to satisfy this case.
Now we will move to the case where x<1. If x<1 , ${{\log }_{x}}0.5>2$ implies $0.5<{{x}^{2}}$ .
$\therefore {{x}^{2}}>0.5$
We know 0.5 can be written as $\dfrac{1}{2}$ . So, our inequality becomes:
${{x}^{2}}>\dfrac{1}{2}$
Now, in case of inequality, ${{x}^{2}}>a$ implies $x\in \left( -\infty ,-\sqrt{a} \right)\cup \left( \sqrt{a},\infty \right)$ . Therefore, for our inequality:
$x\in \left( -\infty ,-\sqrt{\dfrac{1}{2}} \right)\cup \left( \sqrt{\dfrac{1}{2}},\infty \right)$ . But in this case x must be less than 1 and according to the domain of x, x cannot be $(-\infty ,0]\cup \\{1\\}$ . Therefore, the possible values of x satisfying this case are:
$x\in \left( \sqrt{\dfrac{1}{2}},1 \right)$
Also, there is no value of x that satisfies the previous case, where x>1, so all possible values of x are $x\in \left( \sqrt{\dfrac{1}{2}},1 \right)$ . Hence, the answer to the above question is option (a).

Note: Whenever dealing with an inequality be very careful while you multiply, square or perform other operations, as there are cases where the sign of inequality changes. For example: x>y implies $-y>-x$ , i.e. , when both sides of an inequality are multiplied by a negative number, the sign of inequality changes. You could have also answered the question just by eliminating the options, as x can have only positive values and there is only one option in which x is having only positive values.