Question

Simplify:
1) ${2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}}$
2) ${\left( {\dfrac{1}{{{3^3}}}} \right)^7}$
3) $\dfrac{{{{11}^{\dfrac{1}{2}}}}}{{{{11}^{\dfrac{1}{4}}}}}$
4) ${7^{\dfrac{1}{2}}}{.8^{\dfrac{1}{2}}}$

Answer 1

Hint:
A fractional exponent is an alternate way to express the powers and roots together. For example, the following are equivalent.
${81^{\dfrac{1}{2}}} = \sqrt[2]{{81}}$
The numerator denotes the power whereas the denominator denotes the index of the root. If there is no power being applied, write “1” in the numerator as a placeholder, which denotes the index of the root as shown below.
${81^{\dfrac{1}{2}\dfrac{{ \to power}}{{ \to \;\quad root}}}}$
Exponent’s rules and properties

Rule name	Rule
Product rules	\[{a^{\;n}}\; \cdot \;{a^{\;m}}\; = \;{a^{\;n + m}}\]
	\[{a^{\;n}}\; \cdot \;{b^{\;n}}\; = {\text{ }}{(a\; \cdot \;b)^{\;n}}\]
Quotient rules	$$\[\dfrac{{{a^{\;n}}\;}}{{{a^{\;m}}\;}} = \;{a^{\;n}}^{ - m}\]
	\[\dfrac{{{a^{\;n}}}}{{{b^{\;n}}}} = {\text{ }}{\left( {\dfrac{a}{b}\;} \right)^{\;n}}\]
Power rules	\[{\left( {{b^n}} \right)^m}\; = \;{b^n}^{ \cdot m}\]
	\[\sqrt[m]{{\left( {{b^n}} \right)}}{\text{ }} = \;b{\;^{\dfrac{n}{m}}}\]
	\[{b^{\dfrac{1}{n}}}\; = \;\sqrt[n]{b}\]
Negative exponents	\[{b^{ - n}}\; = {\text{ }}\dfrac{1}{{\;{b^n}}}\]

Complete step by step solution:
${2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}}$, for simplifying the expression we will be using the first product rule.
\[{a^{\;n}}\; \cdot \;{a^{\;m}}\; = \;{a^{\;n + m}}\]
$ \Rightarrow {2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}} = {2^{\dfrac{2}{3}}} \times {2^{\dfrac{1}{5}}}$
$ \Rightarrow {2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}} = {2^{\dfrac{2}{3} + \dfrac{1}{5}}}$
Now taking the LCM for the two fractions exponentials
$ \Rightarrow {2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}} = {2^{\dfrac{{10 + 3}}{{15}}}}$
$ \Rightarrow {2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}} = {2^{\dfrac{{13}}{{15}}}}$
Therefore, $ \Rightarrow {2^{\dfrac{2}{3}}}{.2^{\dfrac{1}{5}}} = {2^{\dfrac{{13}}{{15}}}}$
${\left( {\dfrac{1}{{{3^3}}}} \right)^7}$, for simplifying the expression we will be using the negative exponents and first power rule.
\[{b^{ - n}}\; = {\text{ }}\dfrac{1}{{\;{b^n}}}\] and \[{\left( {{b^n}} \right)^m}\; = \;{b^n}^{ \cdot m}\]
$ \Rightarrow {\left( {\dfrac{1}{{{3^3}}}} \right)^7} = {\left( {{3^{ - 3}}} \right)^7}$
$ \Rightarrow {\left( {\dfrac{1}{{{3^3}}}} \right)^7} = {3^{ - 3 \times }}^7$
$ \Rightarrow {\left( {\dfrac{1}{{{3^3}}}} \right)^7} = {3^{ - 21}}$
Therefore, \[ \Rightarrow {\left( {\dfrac{1}{{{3^3}}}} \right)^7} = {3^{ - 21}}\]
$\dfrac{{{{11}^{\dfrac{1}{2}}}}}{{{{11}^{\dfrac{1}{4}}}}}$, for simplifying the expression we will be using the quotient rules.
\[\dfrac{{{a^{\;n}}\;}}{{{a^{\;m}}\;}} = \;{a^{\;n}}^{ - m}\]
$ \Rightarrow \dfrac{{{{11}^{\dfrac{1}{2}}}}}{{{{11}^{\dfrac{1}{4}}}}} = {11^{\left( {\dfrac{1}{2} - \dfrac{1}{4}} \right)}}$
Now taking the LCM for the two fractions exponentials
$ \Rightarrow \dfrac{{{{11}^{\dfrac{1}{2}}}}}{{{{11}^{\dfrac{1}{4}}}}} = {11^{\left( {\dfrac{{2 - 1}}{4}} \right)}}$
$ \Rightarrow \dfrac{{{{11}^{\dfrac{1}{2}}}}}{{{{11}^{\dfrac{1}{4}}}}} = {11^{\dfrac{1}{4}}}$
Therefore, $ \Rightarrow \dfrac{{{{11}^{\dfrac{1}{2}}}}}{{{{11}^{\dfrac{1}{4}}}}} = {11^{\dfrac{1}{4}}}$
${7^{\dfrac{1}{2}}}{.8^{\dfrac{1}{2}}}$, for simplifying the expression we will be using the second product rule.
\[{a^{\;n}}\; \cdot \;{b^{\;n}}\; = {\text{ }}{(a\; \cdot \;b)^{\;n}}\]
$ \Rightarrow {7^{\dfrac{1}{2}}}{.8^{\dfrac{1}{2}}} = {\left( {7 \times 8} \right)^{\dfrac{1}{2}}}$
$ \Rightarrow {7^{\dfrac{1}{2}}}{.8^{\dfrac{1}{2}}} = {56^{\dfrac{1}{2}}}$

Therefore, $ \Rightarrow {7^{\dfrac{1}{2}}}{.8^{\dfrac{1}{2}}} = {\left( {7 \times 8} \right)^{\dfrac{1}{2}}}$

Note:
Please note when it comes to dealing with exponents students often make mistakes. Sometimes, they multiply the base and the exponent. For e.g. ${16^{\dfrac{1}{4}}}$is not equal to 4, it's 2. Then when using the multiplication rule don’t forget it only applies to expressions with the same base. For e.g.: - Four squared times two cubed is not the same as 8 raised to the power two plus three.${4^2} \times {2^3} \ne {8^{2 + 3}}$. The last one is the multiplication rule applies just to the product, not to the sum of two numbers.