Question

working with fractional exponents
\\(\sqrtn{x} = x^{\frac{1}{n}}\\)
the nth root of \\(x\\) is the same as \\(x\\) raised to the \\(\frac{1}{n}\\) power.
complete this expression.
\\(\sqrt4{z}=\square\\)
simplify the following expression completely.
\\(\sqrt13{v^9} = \square\\)
simplify the following.
\\(\sqrt3{27} = \square\\)
simplify.
\\(16^{\frac{1}{4}} = \square\\)

Explanation:

First Sub - Question: $\boldsymbol{\sqrt[4]{z}=}$

Step1: Recall the formula

The formula given is $\sqrt[n]{x}=x^{\frac{1}{n}}$. Here, $n = 4$ and $x=z$.

Step2: Apply the formula

Substitute $n = 4$ and $x = z$ into the formula $\sqrt[n]{x}=x^{\frac{1}{n}}$. So we get $\sqrt[4]{z}=z^{\frac{1}{4}}$.

Step1: Recall the formula

We know that $\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$ (derived from $\sqrt[n]{x}=x^{\frac{1}{n}}$ and $(x^{a})^{b}=x^{ab}$). Here, $n = 13$, $m = 9$ and $x = v$.

Step2: Apply the formula

Substitute $n = 13$, $m = 9$ and $x = v$ into the formula $\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$. So we have $\sqrt[13]{v^{9}}=v^{\frac{9}{13}}$.

Step1: Recall the formula and factor 27

We know that $\sqrt[n]{x}=x^{\frac{1}{n}}$. First, factor 27: $27 = 3\times3\times3=3^{3}$. So $\sqrt[3]{27}=\sqrt[3]{3^{3}}$.

Step2: Apply the formula

Using the formula $\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$, here $n = 3$, $m = 3$ and $x = 3$. So $\sqrt[3]{3^{3}}=3^{\frac{3}{3}}$.

Step3: Simplify the exponent

Simplify $3^{\frac{3}{3}}$, since $\frac{3}{3}=1$, we get $3^{1}=3$.

Answer:

$z^{\frac{1}{4}}$

Second Sub - Question: $\boldsymbol{\sqrt[13]{v^{9}}=}$