How do you simplify $ \left( \dfrac{4{{m}^{4}}{{n}^{3}}{{p}^{3}}}{3{{m}^{2}}{{n}^{2}}{{p}^{4}}} \right) $ and write it using only positive exponents?
Answer
629.4k+ views
Hint: Here, in this question, we have to reduce the terms which have the same base. The solution will be expressed in positive exponents only. If powers become negative, we have to make them positive before reaching the final answer. You should be familiar with the properties of exponents and powers.
Complete step by step answer:
Now, let’s solve the question.
Exponents can be expressed in the form: $ {{a}^{x}} $ which can be read as ‘a’ raise to the power ‘x’. Here ‘a’ is the base and ‘x’ is the power or we can say ‘x’ is an exponent. We also know that the value of ‘a’ should be greater than zero and cannot be equal to one. The value of ‘x’ can be a real number.
There are some important functions for exponents. They are as follows:
$ \begin{align}
& \Rightarrow {{a}^{x}}\times {{a}^{y}}={{a}^{x+y}} \\
& \Rightarrow \dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}} \\
& \Rightarrow {{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}} \\
& \Rightarrow {{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}} \\
& \Rightarrow \dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}} \\
& \Rightarrow {{a}^{0}}=1 \\
& \Rightarrow {{a}^{-x}}=\dfrac{1}{{{a}^{x}}} \\
\end{align} $
Now, write the expression given in question:
$ \Rightarrow \left( \dfrac{4{{m}^{4}}{{n}^{3}}{{p}^{3}}}{3{{m}^{2}}{{n}^{2}}{{p}^{4}}} \right) $
Let’s reduce the terms having the same base. First reduce $ {{m}^{2}} $ with $ {{m}^{4}} $ . We get:
$ \Rightarrow \left( \dfrac{4{{m}^{2}}{{n}^{3}}{{p}^{3}}}{3{{n}^{2}}{{p}^{4}}} \right) $
Next step is to reduce $ {{n}^{2}} $ with $ {{n}^{3}} $ . We get:
$ \Rightarrow \left( \dfrac{4{{m}^{2}}n{{p}^{3}}}{3{{p}^{4}}} \right) $
Now finally we have to reduce $ {{p}^{4}} $ with $ {{p}^{3}} $ . We get:
$ \therefore \left( \dfrac{4{{m}^{2}}n}{3p} \right) $
So, this is our final solution with all the positive exponents.
Note:
Students should note that in the final step there is no need to bring ‘p’ in the numerator as power will become negative and in the question, it is asked to find the solution using positive exponents. One should also know about the multiplicative inverse before solving such questions.
Complete step by step answer:
Now, let’s solve the question.
Exponents can be expressed in the form: $ {{a}^{x}} $ which can be read as ‘a’ raise to the power ‘x’. Here ‘a’ is the base and ‘x’ is the power or we can say ‘x’ is an exponent. We also know that the value of ‘a’ should be greater than zero and cannot be equal to one. The value of ‘x’ can be a real number.
There are some important functions for exponents. They are as follows:
$ \begin{align}
& \Rightarrow {{a}^{x}}\times {{a}^{y}}={{a}^{x+y}} \\
& \Rightarrow \dfrac{{{a}^{x}}}{{{a}^{y}}}={{a}^{x-y}} \\
& \Rightarrow {{\left( {{a}^{x}} \right)}^{y}}={{a}^{xy}} \\
& \Rightarrow {{a}^{x}}\times {{b}^{x}}={{\left( ab \right)}^{x}} \\
& \Rightarrow \dfrac{{{a}^{x}}}{{{b}^{x}}}={{\left( \dfrac{a}{b} \right)}^{x}} \\
& \Rightarrow {{a}^{0}}=1 \\
& \Rightarrow {{a}^{-x}}=\dfrac{1}{{{a}^{x}}} \\
\end{align} $
Now, write the expression given in question:
$ \Rightarrow \left( \dfrac{4{{m}^{4}}{{n}^{3}}{{p}^{3}}}{3{{m}^{2}}{{n}^{2}}{{p}^{4}}} \right) $
Let’s reduce the terms having the same base. First reduce $ {{m}^{2}} $ with $ {{m}^{4}} $ . We get:
$ \Rightarrow \left( \dfrac{4{{m}^{2}}{{n}^{3}}{{p}^{3}}}{3{{n}^{2}}{{p}^{4}}} \right) $
Next step is to reduce $ {{n}^{2}} $ with $ {{n}^{3}} $ . We get:
$ \Rightarrow \left( \dfrac{4{{m}^{2}}n{{p}^{3}}}{3{{p}^{4}}} \right) $
Now finally we have to reduce $ {{p}^{4}} $ with $ {{p}^{3}} $ . We get:
$ \therefore \left( \dfrac{4{{m}^{2}}n}{3p} \right) $
So, this is our final solution with all the positive exponents.
Note:
Students should note that in the final step there is no need to bring ‘p’ in the numerator as power will become negative and in the question, it is asked to find the solution using positive exponents. One should also know about the multiplicative inverse before solving such questions.
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