Question

How do you simplify $\left( \dfrac{4{{a}^{2}}b}{{{a}^{3}}{{b}^{2}}} \right)\left( \dfrac{5{{a}^{2}}b}{2{{b}^{4}}} \right)$ and write it using only positive exponents.

Answer 1

Hint: Now to simplify the given expression we will first open the brackets in the expression and multiply the terms in numerator and denominator. For multiplication we know that ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$ . Now we will use division law which is $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ to further simplify the expression. Now we will convert the negative powers to positive by using the property $\dfrac{1}{{{a}^{n}}}={{a}^{-n}}$ .

Complete step by step solution:
Let us first understand the concept of indices.
Indices are nothing but a number raised to a number or variable.
Hence we write indices as ${{2}^{4}}$ where 4 is raised to 2.
Now in the example ${{2}^{4}}$, 4 is known as power of 2.
Now let us understand the meaning of ${{2}^{4}}$ .
Now the power of a number or variable tells us how many times the number or variable is multiplied by itself. Since here 4 is the power given then we will multiply 2, 4 times.
Hence we have ${{2}^{4}}=2\times 2\times 2\times 2$ .
Now let us first understand three rules of indices.
Now according to the multiplication rule ${{a}^{m}}{{a}^{n}}={{a}^{m+n}}$ , according to division rule we have $\dfrac{{{a}^{m}}}{{{a}^{n}}}={{a}^{m-n}}$ and then the exponent rule states that ${{\left( {{a}^{m}} \right)}^{n}}={{a}^{m\times n}}$ .
Now consider the given expression $\left( \dfrac{4{{a}^{2}}b}{{{a}^{3}}{{b}^{2}}} \right)\left( \dfrac{5{{a}^{2}}b}{2{{b}^{4}}} \right)$
Now first we will first open the brackets and multiply the terms in numerator and denominator by using the multiplication rule. Hence we get,
$\begin{align}
  & \Rightarrow \left( \dfrac{20{{a}^{2+2}}{{b}^{1+1}}}{2{{a}^{3}}{{b}^{2+4}}} \right) \\
 & \Rightarrow \left( \dfrac{20{{a}^{4}}{{b}^{2}}}{2{{a}^{3}}{{b}^{6}}} \right) \\
\end{align}$
Now using the division rule we get,
$\begin{align}
  & \Rightarrow 10{{a}^{4-3}}{{b}^{2-6}} \\
 & \Rightarrow 10a{{b}^{-4}} \\
\end{align}$
Now we want the powers to be positive.
Now we know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ hence using this we get,
$\Rightarrow \dfrac{10a}{{{b}^{4}}}$
Hence the given expression can be written as $\dfrac{10a}{{{b}^{4}}}$ .

Note:
Now note that if we have 0 in the power to any number or variable then the value of the exponent is 1. Hence we have ${{a}^{0}}=1$ . Now consider the term $\dfrac{1}{{{a}^{n}}}$ we can write 1 as ${{a}^{0}}$ hence we get $\dfrac{{{a}^{0}}}{{{a}^{n}}}$ Now using division law we get $\dfrac{{{a}^{0}}}{{{a}^{n}}}={{a}^{0-n}}={{a}^{-n}}$ . Hence we have $\dfrac{1}{{{a}^{n}}}={{a}^{-n}}$