# How do you write 8^(4/3) in radical form?

Aug 7, 2017

See a solution process below:

#### Explanation:

First, rewrite the expression as:

${8}^{\textcolor{red}{4} \times \textcolor{b l u e}{\frac{1}{3}}}$

Next, use this rule of exponents to rewrite the expression again:

${x}^{\textcolor{red}{a} \times \textcolor{b l u e}{b}} = {\left({x}^{\textcolor{red}{a}}\right)}^{\textcolor{b l u e}{b}} 9$

${8}^{\textcolor{red}{4} \times \textcolor{b l u e}{\frac{1}{3}}} = {\left({8}^{\textcolor{red}{4}}\right)}^{\textcolor{b l u e}{\frac{1}{3}}}$

Now, use this rule for exponents and radicals to write this in radical form:

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

${\left({8}^{4}\right)}^{\frac{1}{\textcolor{red}{3}}} = \sqrt[\textcolor{red}{3}]{{8}^{4}}$

If necessary, we can reduce this further by first rewriting the radical as:

$\sqrt[3]{{8}^{3} \cdot 8}$

We can then use this rule for multiplying radicals to simplify the radical:

$\sqrt[n]{\textcolor{red}{a} \cdot \textcolor{b l u e}{b}} = \sqrt[n]{\textcolor{red}{a}} \cdot \sqrt[n]{\textcolor{b l u e}{b}}$

$\sqrt[3]{\textcolor{red}{{8}^{3}} \cdot \textcolor{b l u e}{8}} = \sqrt[3]{\textcolor{red}{{8}^{3}}} \cdot \sqrt[3]{\textcolor{b l u e}{8}} = 8 \sqrt[3]{8}$

Aug 7, 2017

The denominator of the exponent tells us the root and the numerator tells us the power.

#### Explanation:

One way to recall which is which is to think about ${n}^{\frac{1}{5}}$ and ${n}^{5}$ The second is the same as ${x}^{\frac{5}{1}}$ so the numeraotr gives the power and the denominator gives the root.

${8}^{\frac{4}{3}}$

This exponent is a reduced fraction.

If the fraction exponent is already reduced the it doesn't matter what order we use.

${8}^{\frac{4}{3}} = \sqrt[3]{{8}^{4}}$ is the same as ${\sqrt[3]{8}}^{4}$.

In fact, in the second for, if we notice that we know $\sqrt[3]{8} = 2$, then we can quickly simplify further.

${\sqrt[3]{8}}^{4} = {\left(\sqrt[3]{8}\right)}^{4} = {\left(2\right)}^{4} = 16$

I've tried to show the thought process. It would be fine to write just

${\sqrt[3]{8}}^{4} = {2}^{4} = 16$.

Final note

We could also simplify ${8}^{\frac{4}{3}} = \sqrt[3]{{8}^{4}}$

(We'd better be able to. I just said they are the same.)

${8}^{\frac{4}{3}} = \sqrt[3]{{8}^{4}} = \sqrt[3]{{8}^{3} \cdot 8}$

$= \sqrt[3]{{8}^{3}} \sqrt[3]{8}$

$= 8 \sqrt[3]{8}$

$= 8 \cdot 2$

$= 16$