# How do you express t^(-2/7) in radical form?

Jun 3, 2018

See a solution process below:

#### Explanation:

We can rewrite this expression as:

${t}^{- 2 \times \frac{1}{7}}$

Now, we can use this rule for exponents to rewrite the expression again:

${x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}} = {\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}}$

${t}^{\textcolor{red}{- 2} \times \textcolor{b l u e}{\frac{1}{7}}} \implies {\left({t}^{\textcolor{red}{- 2}}\right)}^{\textcolor{b l u e}{\frac{1}{7}}}$

Now, we can use this rule to write the expression in radical form:

${x}^{\frac{1}{\textcolor{red}{n}}} = \sqrt[\textcolor{red}{n}]{x}$

${\left({t}^{-} 2\right)}^{\frac{1}{\textcolor{red}{7}}} \implies \sqrt[\textcolor{red}{7}]{{t}^{-} 2}$

If it is necessary to have no negative exponents we can use this rule of exponents to eliminate the negative exponent:

${x}^{\textcolor{red}{a}} = \frac{1}{x} ^ \textcolor{red}{- a}$

$\sqrt[7]{{t}^{\textcolor{red}{- 2}}} \implies \sqrt[7]{\frac{1}{t} ^ \textcolor{red}{- - 2}} \implies \sqrt[7]{\frac{1}{t} ^ \textcolor{red}{2}}$

Or, because the $n$th root of $1$ is always $1$

$\frac{1}{\sqrt[7]{{t}^{2}}}$