Question

How do you express $\dfrac{{{x}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{3}}}}$ in the radical form.

Answer 1

Hint: Now to solve the expression first we will simplify it by using the property $\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$ . Now we will take the LCM of denominators and subtract the fraction in power of x. Now we know that according to the rule of fractional exponents ${{x}^{\dfrac{p}{q}}}=\sqrt[q]{{{x}^{p}}}$ . Hence we will use this to write the expression in Radical form.

Complete step-by-step solution:
First let us understand the laws of fraction and power.
Now first let us learn product property and quotient property of exponents.
According to product property we have ${{x}^{m}}\times {{x}^{n}}={{x}^{m+n}}$
According to quotient property we have $\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$ .
Now let us understand the rule of fractional exponents.
According to the rule of fractional exponents we can write the term ${{x}^{\dfrac{p}{q}}}$ as $\sqrt[q]{{{x}^{p}}}$ .
Now we can use these properties to simplify the given expression.
Consider the given expression $\dfrac{{{x}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{3}}}}$ .
Now we know that $\dfrac{{{x}^{m}}}{{{x}^{n}}}={{x}^{m-n}}$ Hence using this property we get,
$\Rightarrow \dfrac{{{x}^{\dfrac{1}{2}}}}{{{x}^{\dfrac{1}{3}}}}={{x}^{\left( \dfrac{1}{2}-\dfrac{1}{3} \right)}}$
Now we will take LCM of the denominators to subtract the fractions in power hence we get,
$\begin{align}
  & \Rightarrow {{x}^{\left( \dfrac{3-2}{6} \right)}} \\
 & \Rightarrow {{x}^{\dfrac{1}{6}}} \\
\end{align}$
Now we know that any term of the form ${{x}^{\dfrac{p}{q}}}$ can be written as $\sqrt[q]{{{x}^{p}}}$ hence using this we get,
$\Rightarrow {{x}^{\dfrac{1}{6}}}=\sqrt[6]{x}$
Hence the given term can be expressed in radical form as $\sqrt[6]{x}$

Note: Now note that when we have negative power in numerator we write it as positive power in denominator. Similarly if we have negative power in denominator we write it as positive power in denominator. Hence we have ${{x}^{-p}}=\dfrac{1}{{{x}^{p}}}$ this is because 1 is nothing but ${{x}^{0}}$ and using the property of quotient we get $\dfrac{1}{{{x}^{p}}}=\dfrac{{{x}^{0}}}{{{x}^{p}}}={{x}^{0-p}}={{x}^{-p}}$ .