Question

What is the value of ${{e}^{o}},{{e}^{1}}$ ?

Answer 1

Hint:We first explain the process of exponents and indices. We find the general form. Then we explain the values for ${{e}^{o}},{{e}^{1}}$. We also use the decimal value of $e$. We find the final solution.

Complete step by step answer:
We know the exponent form of the number $a$ with the exponent being $n$ can be expressed as ${{a}^{n}}$. The simplified form of the expression ${{a}^{n}}$ can be written as the multiplied form of number $a$ of n-times.Therefore,
${{a}^{n}}=\underbrace{a\times a\times a\times ....\times a\times a}_{n-times}$
The value of $n$ can be any number belonging to the domain of real numbers.Similarly, the value of $a$ can be any number belonging to the domain of real numbers.

In case the value of $n$ becomes negative, the value of the exponent takes its inverse value.The formula to express the form is,
${{a}^{-n}}=\dfrac{1}{{{a}^{n}}},n\in {{\mathbb{R}}^{+}}$
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 $n$.Let the numbers be ${{a}^{m}}$ and ${{a}^{n}}$. We take multiplication of these numbers.The indices get added. So, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$.The division works in an almost similar way. The indices get subtracted. So, $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$.
We know that $e=2.732$.
We know that ${{a}^{0}}=1,\forall a$.
Therefore, ${{e}^{o}}=1,{{e}^{1}}=e=2.732$.

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 ${{a}^{n}}$, we have ${{a}^{m}}\pm {{a}^{n}}={{a}^{m}}\left( 1\pm {{a}^{n-m}} \right)$.the relation is independent of the values of $m$ and $n$.