
How do you write ${{8}^{4/3}}$ in radical form?
Answer
542.1k+ views
Hint: We know that radicals are also called "roots". They are the "opposite" operation of applying exponents. We can reverse power with a radical, and we can very well reverse a radical with a power. For instance, $2$ when squared gives $4$, and if the square root of $4$ is taken, we get $2$. Similarly, if we take a square of $3$, we get $9$, and on taking the square root of $9$, we get $3$. We can denote these sentences in mathematical form in the following way :
$\begin{align}
& {{2}^{2}}=4,so\sqrt{4}=2 \\
& {{3}^{2}}=9,so\sqrt{9}=3 \\
\end{align}$
To simplify a term containing a square root, we generally " carry out " any number that is present as a "perfect square". We factor the number inside the radical symbol and if any number is present in the form of a square that number is pulled out. For example, $9$ is the square of $3$, so the square root of $9$ contains two copies of the factor $3$. Hence, we can take one $3$ out in the front and leave nothing except $1$ inside the radical symbol.
Complete Step by Step Solution:
We have to write ${{8}^{4/3}}$ in radical form.
Now , we can use the properties of exponents to first simplify ${{8}^{4/3}}$
We know by the property of exponents that ${{a}^{x/y}}$ can be written as
${{a}^{x/y}}={{a}^{x\times 1/y}}$Using this , we get
${{8}^{4/3}}={{8}^{4\times \dfrac{1}{3}}}$
Now , we know ${{({{a}^{b}})}^{c}}={{a}^{(b\times c)}}$
Using this formula , we get
${{8}^{4\times 1/3}}={{({{8}^{4}})}^{\dfrac{1}{3}}}$
According to the rule of radical
${{a}^{1/n}}=\sqrt[n]{a}or\sqrt[n]{a}={{a}^{1/n}}$Thus the radical form of ${{8}^{4/3}}$ is
${{8}^{(4/3)}}={{8}^{(4\times 1/3)}}={{({{8}^{4}})}^{\dfrac{1}{3}}}=\sqrt[3]{{{8}^{4}}}$Now we can also ${{8}^{4}}$as ${{8}^{3+1}}$ and according to the property of exponents ${{a}^{b+c}}={{a}^{b}}.{{a}^{c}}$, we get
$\sqrt[3]{{{8}^{4}}}=\sqrt[3]{{{8}^{3}}.8}=\sqrt[3]{{{8}^{3}}}.\sqrt[3]{8}=8.\sqrt[3]{8}$
which is the final answer.
Note:
There are many properties of exponents that we can use in simplifying or reducing the expression which can make the solution much easier. So, it is noteworthy to keep these points in mind and use the formula as and when required.
$\begin{align}
& {{2}^{2}}=4,so\sqrt{4}=2 \\
& {{3}^{2}}=9,so\sqrt{9}=3 \\
\end{align}$
To simplify a term containing a square root, we generally " carry out " any number that is present as a "perfect square". We factor the number inside the radical symbol and if any number is present in the form of a square that number is pulled out. For example, $9$ is the square of $3$, so the square root of $9$ contains two copies of the factor $3$. Hence, we can take one $3$ out in the front and leave nothing except $1$ inside the radical symbol.
Complete Step by Step Solution:
We have to write ${{8}^{4/3}}$ in radical form.
Now , we can use the properties of exponents to first simplify ${{8}^{4/3}}$
We know by the property of exponents that ${{a}^{x/y}}$ can be written as
${{a}^{x/y}}={{a}^{x\times 1/y}}$Using this , we get
${{8}^{4/3}}={{8}^{4\times \dfrac{1}{3}}}$
Now , we know ${{({{a}^{b}})}^{c}}={{a}^{(b\times c)}}$
Using this formula , we get
${{8}^{4\times 1/3}}={{({{8}^{4}})}^{\dfrac{1}{3}}}$
According to the rule of radical
${{a}^{1/n}}=\sqrt[n]{a}or\sqrt[n]{a}={{a}^{1/n}}$Thus the radical form of ${{8}^{4/3}}$ is
${{8}^{(4/3)}}={{8}^{(4\times 1/3)}}={{({{8}^{4}})}^{\dfrac{1}{3}}}=\sqrt[3]{{{8}^{4}}}$Now we can also ${{8}^{4}}$as ${{8}^{3+1}}$ and according to the property of exponents ${{a}^{b+c}}={{a}^{b}}.{{a}^{c}}$, we get
$\sqrt[3]{{{8}^{4}}}=\sqrt[3]{{{8}^{3}}.8}=\sqrt[3]{{{8}^{3}}}.\sqrt[3]{8}=8.\sqrt[3]{8}$
which is the final answer.
Note:
There are many properties of exponents that we can use in simplifying or reducing the expression which can make the solution much easier. So, it is noteworthy to keep these points in mind and use the formula as and when required.
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