Question

Express 256 as power of 4.

Answer 1

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 number for the formula ${{a}^{n}}=\underbrace{a\times a\times a\times ....\times a\times a}_{n-times}$ and express 256 as power of 4.

Complete step by step answer:
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 number. The value of $a$ can be any number belonging to the domain of real number.
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, ${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$. We also got ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}$.
We find the factorisation of 256 as
\[\begin{align}
  & 2\left| \!{\underline {\,
  256 \,}} \right. \\
 & 2\left| \!{\underline {\,
  128 \,}} \right. \\
 & 2\left| \!{\underline {\,
  64 \,}} \right. \\
 & 2\left| \!{\underline {\,
  32 \,}} \right. \\
 & 2\left| \!{\underline {\,
  16 \,}} \right. \\
 & 2\left| \!{\underline {\,
  8 \,}} \right. \\
 & 2\left| \!{\underline {\,
  4 \,}} \right. \\
 & 2\left| \!{\underline {\,
  2 \,}} \right. \\
 & 1\left| \!{\underline {\,
  1 \,}} \right. \\
\end{align}\]
So, $256=2\times 2\times 2\times 2\times 2\times 2\times 2\times 2={{2}^{8}}$.
We get ${{2}^{8}}={{2}^{2\times 4}}={{\left( {{2}^{2}} \right)}^{4}}={{4}^{4}}$.

So, expressing 256 as a power of 4, we get ${{2}^{8}}$.

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 $n$.