Question

Simplify this expression ${{a}^{m}}\times {{a}^{n}}$.
(a) ${{a}^{m}}+{{a}^{n}}$
(b) ${{a}^{m-n}}$
(c) ${{a}^{m+n}}$
(d) ${{a}^{mn}}$

Answer 1

Hint: Before going to solve this problem, we will discuss the concept of Base and Exponent. When we want to calculate the value ${{2}^{3}}$. Here $2$ is called as base and $3$ is called as exponent and the value of ${{2}^{3}}$ is calculated as ${{2}^{3}}=2\times 2\times 2=8$. From this we can say that the value of ${{a}^{m}}$ where $a$ is the base and $m$ is the exponent is obtained by multiplying the base $a$ with itself $m$(exponent) time. We have some basic exponent rules involving binary operations of bases and exponents. These are the following rules, (${{x}^{a}}{{x}^{b}}={{x}^{a+b}}$; $\dfrac{{{x}^{a}}}{{{x}^{b}}}={{x}^{a-b}}$; ${{\left( {{x}^{a}} \right)}^{b}}={{x}^{ab}}$; ${{\left( xy \right)}^{a}}={{x}^{a}}{{y}^{a}}$.

Complete step-by-step answer:
Let us assume that $a=2$, $m=1$, $n=2$ then the value of ${{a}^{m}}\times {{a}^{n}}$is
$\begin{align}
  & {{a}^{m}}\times {{a}^{n}}={{2}^{1}}\times {{2}^{2}} \\
 & \Rightarrow {{a}^{m}}\times {{a}^{n}}=2\times 4 \\
 & \therefore {{a}^{m}}\times {{a}^{n}}=8 \\
\end{align}$
Now the value of ${{a}^{m}}+{{a}^{n}}$ is
$\begin{align}
  & {{a}^{m}}+{{a}^{n}}={{2}^{1}}+{{2}^{2}} \\
 & \Rightarrow {{a}^{m}}+{{a}^{n}}=2+4 \\
 & \therefore {{a}^{m}}+{{a}^{n}}=6......\left( \text{i} \right) \\
\end{align}$
The value of ${{a}^{m-n}}$ is
$\begin{align}
  & {{a}^{m-n}}={{2}^{1-2}} \\
 & \Rightarrow {{a}^{m-n}}={{2}^{-1}} \\
 & \therefore {{a}^{m-n}}=\dfrac{1}{2}.....\left( \text{ii} \right) \\
\end{align}$
The value of ${{a}^{m+n}}$ is
$\begin{align}
  & {{a}^{m+n}}={{2}^{1+2}} \\
 & \Rightarrow {{a}^{m+n}}={{2}^{3}} \\
 & \therefore {{a}^{m+n}}=8.....\left( \text{iii} \right) \\
\end{align}$
The value of ${{a}^{mn}}$ is
$\begin{align}
  & {{a}^{mn}}={{2}^{1\left( 2 \right)}} \\
 & \Rightarrow {{a}^{mn}}={{2}^{2}} \\
 & \therefore {{a}^{mn}}=4......\left( \text{iv} \right) \\
\end{align}$
From equations $\left( \text{i} \right),\left( \text{ii} \right),\left( \text{iii} \right),\left( \text{iv} \right)$ , we can say that the value of ${{a}^{m}}\times {{a}^{n}}$ is equals to the value of ${{a}^{m+n}}$, hence we can say that
${{a}^{m}}\times {{a}^{n}}={{a}^{m+n}}$

Note: For this kind of problem it’s the better way to assume the values and check for the option. Some of the similar problem that can be solved by this method are
$\begin{align}
  & \dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}} \\
 & {{\left( {{a}^{m}} \right)}^{n}}={{a}^{mn}}
\end{align}$