Question

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

Answer 1

In order to find the solution of the given question that is to find how to simplify ${{\log }_{10}}\left( \dfrac{1}{100} \right)-{{\log }_{10}}\left( \dfrac{1}{1000} \right)$, apply the following identities that are $\dfrac{1}{{{a}^{n}}}={{a}^{-n}}$, ${{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x$ and ${{\log }_{a}}a=1$. Reduce it further to the simplified answer.

Complete step by step solution:
According to the question, given expression in the question is as follows:
${{\log }_{10}}\left( \dfrac{1}{100} \right)-{{\log }_{10}}\left( \dfrac{1}{1000} \right)$
Let us consider
$P={{\log }_{10}}\left( \dfrac{1}{100} \right)-{{\log }_{10}}\left( \dfrac{1}{1000} \right)$
We know that $100={{10}^{2}}$ and $1000={{10}^{3}}$, we can rewrite the above equation as follows:
$\Rightarrow P={{\log }_{10}}\left( \dfrac{1}{{{10}^{2}}} \right)-{{\log }_{10}}\left( \dfrac{1}{{{10}^{3}}} \right)$
Now apply the identity $\dfrac{1}{{{a}^{n}}}={{a}^{-n}}$ on $\dfrac{1}{{{10}^{2}}}$ and $\dfrac{1}{{{10}^{3}}}$, we get:
$\Rightarrow P={{\log }_{10}}\left( {{10}^{-2}} \right)-{{\log }_{10}}\left( {{10}^{-3}} \right)...\left( 1 \right)$
After this to further simplify the above equation apply one of the identities of logarithm that is ${{\log }_{a}}{{x}^{n}}=n{{\log }_{a}}x$ on ${{\log }_{10}}\left( {{10}^{-2}} \right)$ and ${{\log }_{10}}\left( {{10}^{-3}} \right)$ that is we can write:
$\Rightarrow {{\log }_{10}}\left( {{10}^{-2}} \right)=\left( -2 \right){{\log }_{10}}\left( 10 \right)...\left( 2 \right)$
$\Rightarrow {{\log }_{10}}\left( {{10}^{-3}} \right)=\left( -3 \right){{\log }_{10}}\left( 10 \right)...\left( 3 \right)$
Now put the equation $\left( 2 \right)$ and $\left( 3 \right)$ in the equation $\left( 1 \right)$, we get:
$\Rightarrow P=\left( -2 \right){{\log }_{10}}\left( 10 \right)-\left( -3 \right){{\log }_{10}}\left( 10 \right)$
To simplify it further, solve the bracket by using one of the concepts of the number system that is two ‘minuses’ makes plus in the above equation, we get:
$\Rightarrow P=-2{{\log }_{10}}\left( 10 \right)+3{{\log }_{10}}\left( 10 \right)$
Now apply another identity of logarithm that is ${{\log }_{a}}a=1$ on ${{\log }_{10}}\left( 10 \right)$ in the above equation, we get:
$\Rightarrow P=-2\left( 1 \right)+3\left( 1 \right)$
Solve the above equation by removing brackets with the help of multiplication, we get:
$\Rightarrow P=-2+3$
To get the final answer, solve the above equation with help of subtraction, we get:
$\Rightarrow P=1$
$\Rightarrow {{\log }_{10}}\left( \dfrac{1}{100} \right)-{{\log }_{10}}\left( \dfrac{1}{1000} \right)=1$

Therefore, after the simplification of the given equation ${{\log }_{10}}\left( \dfrac{1}{100} \right)-{{\log }_{10}}\left( \dfrac{1}{1000} \right)$ we get $1$ as the answer.

Note: Students can go wrong while solving the given question by ignoring the given data that is important to find a solution, using a data which is different from the given data. adding irrelevant or extraneous data, applying a theorem or definition outside its condition, applying a distributive property to a non-distributive function or operation, incorrectly citing a definition, theorem, rule or formula unverified identities of logarithm, error in examining the final result, technical error that is the error in calculation due to carelessness and the error in manipulating algebraic symbol or operation.