Question

If $\dfrac{1}{2} \leqslant {\log _{0.1}}x \leqslant 2$ then:
A. maximum value of $x$ is $\dfrac{1}{{\sqrt {10} }}$
B. $x$ lies between $\dfrac{1}{{100}}$ and $\dfrac{1}{{\sqrt {10} }}$
C. minimum value of $x$ is $\dfrac{1}{{10}}$
D. minimum value of $x$ is $\dfrac{1}{{100}}$
E. maximum value of $x$ is $\dfrac{1}{{100}}$

Answer 1

Hint: To find out the range of the value of $x$ we need to solve the given equation two times.
Firstly we have to solve taking ${\log _{0.1}}x \leqslant 2$ to get the maximum range of value and
Also secondly we need to solve ${\log _{0.1}}x \geqslant \dfrac{1}{2}$ to get the minimum range.
Finally we get the required answer.

Complete step-by-step answer:
At first we will find out the maximum range of value of $x$
So we have to solve ${\log _{0.1}}x \leqslant 2$
We can also write in this way that, ${\log _{0.1}}x \leqslant {\log _{0.1}}(0.1)_{}^2$
Therefore in this case the inequality sign changes because the base of $\log $ is less than $1$ so we can write-
$x \geqslant (0.1)_{}^2$
After removing decimal in the right side we get-
$x \geqslant \dfrac{1}{{100}}$…..(i)
Therefore the minimum value of x is within $\dfrac{1}{{100}}$
Now we have to solve ${\log _{0.1}}x \geqslant \dfrac{1}{2}$ to find out the minimum value of $x$
We can write the equation in this way, ${\log _{0.1}}x \geqslant {\log _{0.1}}(0.1)_{}^{\dfrac{1}{2}}$
Therefore in this case also the inequality sign changes as the base value is less than $1$ so we can write-
$x \leqslant (0.1)_{}^{\dfrac{1}{2}}$
After removing the decimal and doing square root since the power of $0.1$ is $\dfrac{1}{2}$ we get-
$x \leqslant \dfrac{1}{{\sqrt {10} }}$….(ii)
And, the maximum value of x is $\dfrac{1}{{\sqrt {10} }}$
So from equation (i) and (ii) we can say that the value of x ranges from $\dfrac{1}{{100}}$ to $\dfrac{1}{{\sqrt {10} }}$
Thus $x$ lies between $\dfrac{1}{{100}}$ and $\dfrac{1}{{\sqrt {10} }}$
Thus the minimum value is $\dfrac{1}{{100}}$ and the maximum value is $\dfrac{1}{{\sqrt {10} }}$.

Hence the correct option is A,B and D.

Note: It is to be noted that when the base value of log is less than one the inequality sign gets changed. Logarithm is the inverse function to the exponential function. Logarithm is the power to which a number must be raised to get some other number.