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How do you simplify $\dfrac{{{n}^{2}}\times {{n}^{3}}}{{{n}^{4}}}$?

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Last updated date: 13th Jun 2024
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Answer
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Hint: We first explain the process of exponents and indices. We find the general form. Then we explain the different binary operations on exponents. Finally, we find the indices value for $\dfrac{{{n}^{2}}\times {{n}^{3}}}{{{n}^{4}}}$ and express it in that form of power of $n$.

Complete step-by-step solution:
We know the exponent form of the number $a$ with the exponent being $p$ can be expressed as ${{a}^{p}}$.
The simplified form of the expression ${{a}^{p}}$ can be written as the multiplied form of number $a$ of p-times.
Therefore, ${{a}^{p}}=\underbrace{a\times a\times a\times ....\times a\times a}_{p-times}$.
The value of $p$ can be any number belonging to the domain of real number.
Similarly, the value of $a$ can be any number belonging to the domain of real numbers.
In case the value of $p$ becomes negative, the value of the exponent takes its inverse value.
The formula to express the form is ${{a}^{-p}}=\dfrac{1}{{{a}^{p}}},p\in {{\mathbb{R}}^{+}}$.
The multiplication of these exponents works as the addition of those indices.
For example, we take two exponential expressions where the exponents are $m$ and $p$.
Let the numbers be ${{a}^{m}}$ and ${{a}^{p}}$. We take multiplication of these numbers.
The indices get added. So, ${{a}^{m}}\times {{a}^{p}}={{a}^{m+p}}$.
The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{p}}}={{a}^{m-p}}$.
For our given expression, we need to express $\dfrac{{{n}^{2}}\times {{n}^{3}}}{{{n}^{4}}}$ in its exponential way.
First, we complete the numerator part of multiplication where ${{n}^{2}}\times {{n}^{3}}={{n}^{2+3}}={{n}^{5}}$.
The equation becomes $\dfrac{{{n}^{2}}\times {{n}^{3}}}{{{n}^{4}}}=\dfrac{{{n}^{5}}}{{{n}^{4}}}$. Now we do the division.
We have $\dfrac{{{n}^{5}}}{{{n}^{4}}}={{n}^{5-4}}={{n}^{1}}=n$. Therefore, simplified form of $\dfrac{{{n}^{2}}\times {{n}^{3}}}{{{n}^{4}}}$ is $n$.

Note: The addition and subtraction for exponents works for taking common terms out depending on the values of the indices.
For numbers ${{a}^{m}}$ and ${{a}^{p}}$, we have ${{a}^{m}}\pm {{a}^{p}}={{a}^{m}}\left( 1\pm {{a}^{p-m}} \right)$.the relation is independent of the values of $m$ and $p$.