Question

How do you rewrite ${{y}^{-5}}$ using a positive exponent ?

Answer 1

Hint: Here $y$ has a negative power which is $-5$ . To convert it, let’s make use of simple rules of exponents. From our previous knowledge of exponents, we know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ . Let’s use this and solve this question.

Complete step by step answer:
Let’s make use of some random variable. Let it be $x$.
Please note that $x$ is nowhere mentioned in the question. But we using it only for the sake of simplifying our question.
Let $x={{y}^{-5.}}$.
As we have already mentioned in the hint that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ . Let’s make use of it . Upon making use of it , we get the following :
\[\begin{align}
  & \Rightarrow x={{y}^{-5}} \\
 & \Rightarrow x=\dfrac{1}{{{y}^{5}}} \\
\end{align}\]
Now let’s cross multiply . Upon doing so , we get the following :
$\Rightarrow {{y}^{5}}=\dfrac{1}{x}$
Now, let’s take ${{5}^{th}}$root of power on both sides .
It’s almost like taking square root on both side. While applying square root on both sides we multiply the powers present to the terms with $\dfrac{1}{2}$ .
In the same way , when we apply the ${{5}^{th}}$ root on both sides we multiply the powers present to the terms with $\dfrac{1}{5}$ .
Upon doing so , we get the following :
$\Rightarrow {{\left( {{y}^{5}} \right)}^{\dfrac{1}{5}}}={{\left( \dfrac{1}{x} \right)}^{\dfrac{1}{5}}}$
Let us apply here another rule of exponents. This rule states the following :
${{\left( {{a}^{n}} \right)}^{m}}={{a}^{n\times m}}$
Upon using this formula , we get the following :
$\begin{align}
  & \Rightarrow {{y}^{\times \dfrac{1}{{}}}}={{\left( \dfrac{1}{x} \right)}^{\dfrac{1}{5}}} \\
 & \Rightarrow y={{\left( \dfrac{1}{x} \right)}^{\dfrac{1}{5}}} \\
\end{align}$

$\therefore $ Hence we presented ${{y}^{-5}}$ in the form of $y={{\left( \dfrac{1}{x} \right)}^{\dfrac{1}{5}}}$ i.e in a positive power by making use of a random variable $x.$

Note: It is important to know all the rules of exponents so that we can solve the sum easily. While we are solving these kinds of sums , it is very important to clearly mention that the other variable we are using is nowhere related to the question and it is just used as a means of arriving at the answer.