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: First, apply the base change rule law. Then, substitute the value of the log. After that, multiply the equation by -1, which will change the sign of inequalities. By definition of the logarithm, find the value of x and simplify it.

Formula used: The base change law of log is,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$

Complete step-by-step answer:
Given: - $\dfrac{1}{2} \leqslant {\log _{0.1}}x \leqslant 2$
Apply the base change rule law of logarithm,
$\dfrac{1}{2} \leqslant \dfrac{{\log x}}{{\log 0.1}} \leqslant 2$
Substitute the value $\log 0.1 = - 1$ in the equation,
$\dfrac{1}{2} \leqslant \dfrac{{\log x}}{{ - 1}} \leqslant 2$
Multiply the equation by -1. By multiplying with -1, the sign of inequalities will change,
$ - \dfrac{1}{2} \geqslant \log x \geqslant - 2$
By definition of logarithm, if $y = \log x$. Then $x = {10^y}$.
Apply this formula in the equation,
${10^{ - \dfrac{1}{2}}} \geqslant x \geqslant {10^{ - 2}}$
As, we know that ${x^{ - y}} = \dfrac{1}{{{x^y}}}$, then,
$\dfrac{1}{{{{10}^{\dfrac{1}{2}}}}} \geqslant x \geqslant \dfrac{1}{{{{10}^2}}}$
Thus,
$\dfrac{1}{{\sqrt {10} }} \geqslant x \geqslant \dfrac{1}{{100}}$
It shows that the value of x lies between $\dfrac{1}{{100}}$ and $\dfrac{1}{{\sqrt {10} }}$. Also, the minimum value of x is $\dfrac{1}{{100}}$ and the maximum value of x is $\dfrac{1}{{\sqrt {10} }}$.

Hence, option (A), (B), and (D) are correct answers.

Note: The students might make mistakes by not changing the sign of inequalities when multiplied by -1.
$ - \dfrac{1}{2} \geqslant \log x \geqslant - 2$
This will lead to wrong calculations and also give the wrong answer.
Logarithms are the opposite of exponentials, just as the opposite of addition is subtraction and the opposite of multiplication is division.
In other words, a logarithm is essentially an exponent that is written in a particular manner.
Logarithms can make multiplication and division of large numbers easier, because adding logarithms is the same as multiplying, and subtracting logarithms is the same as dividing.
The rules of logarithm are:-
Change of base rule law,
${\log _y}x = \dfrac{{\log x}}{{\log y}}$
Product rule law,
$\log xy = \log x + \log y$
Quotient rule law,
$\log \dfrac{x}{y} = \log x - \log y$
Power rule law,
$\log {x^y} = y\log x$