Exercise 1.5
Maths - Exercise
- Find:
-
- \(64^\frac{1}{2}\)Solution
\[ \begin{aligned} \left( 64\right) ^{\frac{1}{2}}\\ &=\left( 8\times 8\right)^{\frac{1}{2}}\\ &=\left( 8^{2}\right)^{\frac{1}{2}}\end{aligned}\\\\ \color{blue}{\text{Formula: }\left( a^{m}\right)^{n}=a^{mn}}\\\\ \begin{aligned}&=\left( 8\right) ^{\frac{2\times 1}{2}}\\&=8 \end{aligned} \]
- \(64^\frac{1}{2}\)Solution
- \(32^\frac{1}{5}\)Solution \[ \begin{aligned}\left( 32\right) ^{\frac{1}{5}}&=\left( 2\times 2\times 2\times 2\times 2\right) ^{\frac{1}{5}}\\ &=\left( 2^{5}\right) ^{\frac{1}{5}}\end{aligned}\\ \\\color{blue}{\text{Formula: }\left( a^{m}\right) ^{n}=a^{mn}}\\\\\begin{aligned} &=2^{5\cdot \frac{1}{5}}\\&=2 \end{aligned} \]
- \(125^\frac{1}{3}\)Solution \[ \require{cancel} \begin{aligned} 125^\frac{1}{3}&=\left(5\times5\times5\right)^\frac{1}{3}\end{aligned}\\\\ \color{blue}{\text{Using Formula: }\left(a^m\right)^n=a^{mn}}\\\\ \begin{aligned}&=\left(5^3\right)^\frac{1}{3}\\&=5^{3.\frac{1}{3}}\\&=5^{\cancelto{1}{3}.\frac{1}{\cancelto{1}3}}\\&=5 \end{aligned} \]
-
- Find:
- \(9^\frac{3}{2}\)Solution \[ \require{cancel} \begin{aligned}\left( 9\right) ^{\frac{3}{2}}&=\left( 3\times 3\right) ^{\frac{3}{2}}\\ &=\left( 3^{2}\right) ^{3/2}\end{aligned}\\\\\color{blue}{\text{Formula: }\left( a^{m}\right) ^{n}=a^{mn}}\\\\ \begin{aligned}&=\left( (3) ^\left({\dfrac{2\cdot 3}{2}}\right)\right)\\\\ &=\left( (3) ^\left({\dfrac{\cancelto{1}2\cdot 3}{\cancelto{1}2}}\right)\right)\\\\ &=\left( 3\right) ^{3}\\ &=27\end{aligned} \]
- \(32^\frac{2}{5}\)Solution \[ \require{cancel} \begin{aligned} \left( 32\right) ^{2/5}&=\left( 2\times 2\times 2\times 2\times 2\right)^{2/5}\end{aligned} \\\\\color{blue}{\text{Formula: }\left( a^{m}\right) ^{n}=a^{mn}}\\\\ \begin{aligned}&=\left( 2^{5}\right) ^{2/5}\\&=\left( 2\right) ^{5\cdot \dfrac{2}{5}}\\ &=\left( (2) ^{\cancelto{1}5\cdot \dfrac{2}{\cancelto{1}5}}\right)\\ &=2^{2}\\ &=4 \end{aligned} \]
- \(16^\frac{3}{4}\)Solution \[ \require{cancel} \begin{aligned} 16^{3/4}&=\left( 2\times 2\times 2\times 2\right) ^{3/4}\end{aligned}\\\\\color{blue}{\text{Formula: }\left( a^{m}\right)^{n}=a^{mn}}\\\\ \begin{aligned}&=\left( 2^4\right) ^{3/4}\\ &=\left(2^{4}\right) ^{3/4}\\ &=(2)^\left({\cancelto{1}4\cdot \dfrac{3}{\cancelto{1}4}}\right)\\&=2^{3}\\&=8 \end{aligned} \]
- \(125^\frac{-1}{3}\)Solution \[ \require{cancel} \begin{aligned} \left( 125\right) ^{\frac{-1}{3}}&=\dfrac{1}{\left( 125\right) ^{1/3}}\end{aligned}\\\\\color{blue}{\text{Formula: } a^{-m} =\frac{1}{a^m}}\\\\ \begin{aligned}&=\dfrac{1}{\left( 5\times 5\times 5\right) ^{1/3}}\\ &=\frac{1}{\left( 5^{3}\right) ^\left(\frac{1}{3}\right)}\\ &=\frac{1}{5^{3\cdot \frac{1}{3}}}\end{aligned}\\\\ \color{blue}{\text{Formula: } \left(a^m\right)^n=a^{mn}}\\\\ \begin{aligned}&=\frac{1}{(5)^\left({\cancelto{1}3\cdot \frac{1}{\cancelto{1}3}}\right)}\\ &=\frac{1}{\left(5^\left(1\right)\right)}\\&=\frac{1}{5} \end{aligned} \]
- Simplify:
- \(2^\frac{2}{3}.2^\frac{1}{5}\)Solution \[ \begin{aligned} 2^{\frac{2}{3}}\cdot 2^{\frac{1}{5}}&=\left( 2\right) ^{\frac{2}{3}+\frac{1}{5}}\end{aligned}\\\\\color{blue}{\text{Formula: }a^{m}\cdot a^{n}=a^{m+n}}\\\\\begin{aligned} &=\left( 2\right) ^{\frac{10+3}{15}}\\&=2^{\frac{13}{15}} \end{aligned} \]
- \(\left(\frac{1}{3^3}\right)^7\)Solution \[ \begin{aligned} \left( \dfrac{1}{3^{3}}\right) ^{7}&=\dfrac{1^7}{\left( 3^{3}\right) ^{7}}\end{aligned}\\\\\color{blue}{\text{Formula: }\left( \dfrac{a}{b}\right) ^{m}=\dfrac{a^{m}}{b^{m}}} \\\\\begin{aligned}&=\frac{1}{3^{21}}\\&=3^{-21}\\ \end{aligned} \]
- \(\dfrac{11^\frac{1}{2}}{11^\frac{1}{4}}\)Solution \[ \begin{aligned} \dfrac{11^{\frac{1}{2}}}{11^{\frac{1}{4}}}&=11^\left({\frac{1}{2}-\frac{1}{4}}\right) \end{aligned}\\\\\color{blue}{\text{Formula: }\dfrac{a^{m}}{a^{n}}=a^{m-n}}\\\\ \begin{aligned}&=11^\left({\frac{2-1}{4}}\right)\\ &=11^\frac{1}{4} \end{aligned} \]
- \(7^\frac{1}{2}.8^\frac{1}{2}\)Solution \[ \require{cancel} \begin{aligned} 7^{\frac{1}{2}}\cdot 8^{\frac{1}{2}}&= 7^\frac{1}{2} \cdot(2\times 2\times2 )^{\frac{1}{2}}\\ &=7^\frac{1}{2}\cdot(2.2^2)^\frac{1}{2}\\&=7^\frac{1}{2}\cdot2^\frac{1}{2}\cdot2^{2\cdot\frac{1}{2}}\\ &=7^\frac{1}{2}\cdot2^\frac{1}{2}\cdot2^{\cancelto{1}2\cdot\frac{1}{\cancelto{1}2}}\\ &=2\cdot7^\frac{1}{2}\cdot2^\frac{1}{2}\\ &=2(14)^\frac{1}{2} \end{aligned} \]
