Question

hw 1 - algebra for calculus section 0.1: problem 9 (1 point) write the following expressions as $x^{p}$ for some constant $p$. $sqrt{x}=square$ $sqrt3{x}=square$ $sqrt{x^{3}}=square$ $sqrt3{x^{2}}=square$ $\frac{1}{sqrt{x}}=square$ $\frac{1}{sqrt3{x}}=square$

Explanation:

Step1: Recall radical - exponent conversion

The square - root $\sqrt{x}$ can be written as $x^{\frac{1}{2}}$ since $\sqrt[n]{x}=x^{\frac{1}{n}}$, here $n = 2$.

Step2: For cube - root

The cube - root $\sqrt[3]{x}$ can be written as $x^{\frac{1}{3}}$ as $\sqrt[n]{x}=x^{\frac{1}{n}}$ with $n = 3$.

Step3: Simplify $\sqrt{x^{3}}$

Using the rule $\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$, for $\sqrt{x^{3}}$ where $n = 2$ and $m = 3$, we get $x^{\frac{3}{2}}$.

Step4: Simplify $\sqrt[3]{x^{2}}$

Using the rule $\sqrt[n]{x^{m}}=x^{\frac{m}{n}}$, for $\sqrt[3]{x^{2}}$ with $n = 3$ and $m = 2$, we have $x^{\frac{2}{3}}$.

Step5: Simplify $\frac{1}{\sqrt{x}}$

We know that $\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$ because $\frac{1}{x^{a}}=x^{-a}$ and $\sqrt{x}=x^{\frac{1}{2}}$.

Step6: Simplify $\frac{1}{\sqrt[3]{x}}$

Since $\sqrt[3]{x}=x^{\frac{1}{3}}$ and $\frac{1}{x^{a}}=x^{-a}$, we get $x^{-\frac{1}{3}}$.

Answer:

$\sqrt{x}=x^{\frac{1}{2}}$
$\sqrt[3]{x}=x^{\frac{1}{3}}$
$\sqrt{x^{3}}=x^{\frac{3}{2}}$
$\sqrt[3]{x^{2}}=x^{\frac{2}{3}}$
$\frac{1}{\sqrt{x}}=x^{-\frac{1}{2}}$
$\frac{1}{\sqrt[3]{x}}=x^{-\frac{1}{3}}$