# How do you simplify (y^6/y) ^-2 using only positive exponents?

Aug 18, 2016

$\frac{1}{y} ^ 10$

#### Explanation:

${\left(\frac{\cancel{y} \times {y}^{5}}{\cancel{y}}\right)}^{- 2}$

$\frac{1}{{\left({y}^{5}\right)}^{2}}$

$\frac{1}{y} ^ 10$

Aug 18, 2016

$= \frac{1}{y} ^ 10$

#### Explanation:

${\left({y}^{6} / y\right)}^{-} 2$

$= {\left({y}^{5}\right)}^{-} 2$

$= {y}^{-} 10$

$= \frac{1}{y} ^ 10$

Aug 18, 2016

$\frac{1}{y} ^ 10$

#### Explanation:

first observe that ${y}^{{x}^{n}} = {y}^{x n}$ and $\left({y}^{n} / {y}^{1}\right) = {y}^{n - 1}$so

${\left({y}^{6} / y\right)}^{-} 2 = {y}^{-} 10$

when an exponent is negative this is the same as placing the positive version under 1

${y}^{-} 10 = \frac{1}{y} ^ 10$.

Aug 24, 2016

$\frac{1}{y} ^ 10$

#### Explanation:

There is a useful variation of one of the laws of indices.

Note: $\textcolor{b l u e}{{\left(\frac{a}{b}\right)}^{-} c = {\left(\frac{b}{a}\right)}^{+ c}}$

${\left({y}^{6} / y\right)}^{\textcolor{red}{- 2}} = {\left(\frac{y}{y} ^ 6\right)}^{\textcolor{red}{2}}$

There are no longer negative indices, simplify as normal.

=${\left(\frac{1}{y} ^ 5\right)}^{2}$

= $\frac{1}{y} ^ 10$