Question

How do you simplify $${\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right)$$?

Answer 1

Hint: A logarithm is an exponent which indicates to what power a base must be raised to produce a given number.
 $$y = {b^x}$$exponential form,
$$x = {\log _b}y$$ logarithmic function, where $$x$$ is the logarithm of $$y$$ to the base $$b$$, and $${\log _b}y$$ is the power to which we have to raise $$b$$ to get $$y$$, we are expressing $$x$$ in terms of $$y$$.
 Now the given question can solved by using properties of logarithms i.e.,$$\log x - \log y = \log \left( {\dfrac{x}{y}} \right)$$,and $${\log _a}a = 1$$ solve the expression to get the required result.

Complete step-by-step answer:
We know that logarithm is the power to which a number must be raised in order to get some other number, and the base unit is the number being raised to a power, For example, the base ten logarithm of 1000 is 3, because ten raised to the power of two is 100 , because$${10^3} = 1000$$. In general, you write log followed by the base number as a subscript. The most common logarithms are base 10 logarithms and natural logarithms; they have special notations. A base ten log is written as$$\log $$, and we use different base unit but most common logarithms are base 10 logarithms.
Now given equation is$${\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right)$$,
Now using logarithmic property$$\log x - \log y = \log \left( {\dfrac{x}{y}} \right)$$, then the equation becomes,
$$ \Rightarrow {\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right) = {\log _{10}}\left( {\dfrac{{\dfrac{1}{{100}}}}{{\dfrac{1}{{1000}}}}} \right)$$,
Now simplifying the right hand side, we get,
$$ \Rightarrow {\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right) = {\log _{10}}\left( {\dfrac{1}{{\left( {\dfrac{1}{{10}}} \right)}}} \right)$$,
Now taking the denominator of the denominator to the numerator we get,
$$ \Rightarrow {\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right) = {\log _{10}}\left( {10} \right)$$,
By using identity$${\log _a}a = 1$$, we get,
$$ \Rightarrow {\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right) = 1$$.
The value of$${\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right)$$is equal to$$1$$.
Final Answer:

$$\therefore $$ The value of$${\log _{10}}\left( {\dfrac{1}{{100}}} \right) - {\log _{10}}\left( {\dfrac{1}{{1000}}} \right)$$ is equal to $$1$$.

Note:
A logarithm is a mathematical operation that determines how many times a certain number, called the base, is multiplied by itself to reach another number, in these types of questions, we use logarithmic properties and formulas, and some of useful formulas are:
 $${\log _a}xy = {\log _a}x + {\log _a}y$$,
$$\log x - \log y = \log \left( {\dfrac{x}{y}} \right)$$
$${\log _a}{x^n} = n{\log _a}x$$,
$${\log _a}b = \dfrac{{{{\log }_e}b}}{{{{\log }_e}a}}$$,
$${\log _{\dfrac{1}{a}}}b = - {\log _a}b$$,
$${\log _a}a = 1$$,
$${\log _{{a^x}}}b = \dfrac{1}{x}{\log _a}b$$