Question

How do you simplify ${{\left( \dfrac{{{r}^{2}}{{t}^{-3}}}{{{r}^{-3}}{{t}^{5}}} \right)}^{-8}}$ and write it only using positive exponents?

Answer 1

Hint: To simplify ${{\left( \dfrac{{{r}^{2}}{{t}^{-3}}}{{{r}^{-3}}{{t}^{5}}} \right)}^{-8}}$, we are going to use the exponent property which states that when the base is same and the bases are written in the quotient form then the exponents will behave as: $\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$ . We will use this property for “r and t” separately and then rearrange them in such a way so that only positive power will come.

Complete step by step solution:
The expression given in the above problem which we have to simplify is as follows:
${{\left( \dfrac{{{r}^{2}}{{t}^{-3}}}{{{r}^{-3}}{{t}^{5}}} \right)}^{-8}}$
There is a property when the two numbers are same but having different powers and are written in the division form then:
$\dfrac{{{a}^{b}}}{{{a}^{c}}}={{a}^{b-c}}$
Using the above property in r and t in the above expression we get,
${{\left( \dfrac{{{r}^{2}}{{t}^{-3}}}{{{r}^{-3}}{{t}^{5}}} \right)}^{-8}}$
First of all, we will split r and t in the following way:
$\begin{align}
  & {{\left( \dfrac{{{r}^{2}}}{{{r}^{-3}}}\times \dfrac{{{t}^{-3}}}{{{t}^{5}}} \right)}^{-8}} \\
 & ={{\left( {{r}^{2-\left( -3 \right)}}\times {{t}^{-3-5}} \right)}^{-8}} \\
\end{align}$
$\begin{align}
  & ={{\left( {{r}^{2+3}}\times {{t}^{-8}} \right)}^{-8}} \\
 & ={{\left( {{r}^{5}}\times {{t}^{-8}} \right)}^{-8}} \\
\end{align}$
Multiplying the power -8 with the power of r (which is 5) and multiplying the power of -8 with the power of t (which is -8) we get,
$\begin{align}
  & ={{r}^{5\times -8}}\times {{t}^{-8\times -8}} \\
 & ={{r}^{-40}}\times {{t}^{64}} \\
\end{align}$
Now, we need only positive exponents so we can write ${{r}^{-40}}=\dfrac{1}{{{r}^{40}}}$ in the above.
$\dfrac{1}{{{r}^{40}}}\times {{t}^{64}}$
Writing the above expression in compact form we get,
$\Rightarrow \dfrac{{{t}^{64}}}{{{r}^{40}}}$

Hence, we have simplified the given expression into $\dfrac{{{t}^{64}}}{{{r}^{40}}}$

Note: There is another way in which we can write the above solution. The final answer in the above solution is as follows:
$\dfrac{{{t}^{64}}}{{{r}^{40}}}$
Now, as you can see that 8 is common in both the powers of the above expression so we can take the power 8 as outside this fraction and then the above expression will look like:
$={{\left( \dfrac{{{t}^{8}}}{{{r}^{5}}} \right)}^{8}}$
This is also the way in which we can write the above final answer. Sometimes, in the multiple choice questions, a different form of the answer is given so there you can use it.