Question

How do you simplify and write the given term \[6{{y}^{-4}}\] with only positive exponents?

Answer 1

Hint: We start solving the problem by equating the given term to a variable. We then make use of the law of exponents that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$ to proceed through the problem. We then make use of the fact that $6=2\times 3$ in the obtained result to get the simplified form of the given term which is the required answer for the given problem.

Complete step-by-step answer:
According to the problem, we are asked to simplify the given term \[6{{y}^{-4}}\] and write it using only positive exponents.
Let us assume $d=6{{y}^{-4}}$ ---(1).
From the laws of exponents, we know that ${{a}^{-n}}=\dfrac{1}{{{a}^{n}}}$. Let us use this result in equation (1).
$\Rightarrow d=\dfrac{6}{{{y}^{4}}}$ ---(2).
We know that $6=2\times 3$. Let us use this result in equation (2).
$\Rightarrow d=\dfrac{2\times 3}{{{y}^{4}}}$.
So, we have found the required simplified form of the given term \[6{{y}^{-4}}\] in positive exponents as $\dfrac{2\times 3}{{{y}^{4}}}$.
$\therefore $ The required simplified form of the given term \[6{{y}^{-4}}\] in positive exponents is $\dfrac{2\times 3}{{{y}^{4}}}$.

Note: Whenever we get this type of problems, we try to make use of the laws of exponents to get the required answer. We should keep in mind that the exponents for variable y should be positive in the final result. We can also report the obtained answer as $\dfrac{{{2}^{1}}\times {{3}^{1}}}{{{y}^{4}}}$, as we know that ${{2}^{1}}=2$ and ${{3}^{1}}=3$. Similarly, we can expect problems to find the simplified form of the given term ${{\left( \dfrac{16{{y}^{6}}}{{{y}^{-2}}} \right)}^{\dfrac{1}{4}}}$.