# 1.2.1 Indices, PT3 Practice

 
 
1.2.1 Indices, PT3 Practice
 
Question 1:
 
(a) Simplify: a4 ÷ a7
 
(b)   Evaluate:   ${\left({2}^{4}\right)}^{\frac{1}{2}}×{3}^{\frac{1}{2}}×{12}^{\frac{1}{2}}$  
 
 

 Solution:
 
(a) a4 ÷ a7 = a4-7 = a-3
 
 

 (b)
    $\begin{array}{l}{\left({2}^{4}\right)}^{\frac{1}{2}}×{3}^{\frac{1}{2}}×{12}^{\frac{1}{2}}={2}^{2}×{3}^{\frac{1}{2}}×{\left(4×3\right)}^{\frac{1}{2}}\\ \text{}={2}^{2}×{3}^{\frac{1}{2}}×{\left({2}^{2}×3\right)}^{\frac{1}{2}}\\ \text{}={2}^{2}×{3}^{\frac{1}{2}}×2×{3}^{\frac{1}{2}}\\ \text{}={2}^{3}×3\\ \text{}=24\end{array}$  
 
 
 
 
 
Question 2:
 
(a) Simplify: p3 ÷ p-5
 
(b) Evaluate:   ${10}^{\frac{1}{2}}×{5}^{-\frac{1}{2}}×{\left({2}^{\frac{1}{2}}\right)}^{5}$  
 
 

 Solution:
 
(a) p3 ÷ p-5 = p3-(-5) = p3+5 = p8
 
 
(b)
    $\begin{array}{l}{10}^{\frac{1}{2}}×{5}^{-\frac{1}{2}}×{\left({2}^{\frac{1}{2}}\right)}^{5}\\ ={\left(2×5\right)}^{\frac{1}{2}}×{5}^{-\frac{1}{2}}×{2}^{\frac{5}{2}}\\ ={2}^{\frac{1}{2}}×{5}^{\frac{1}{2}}×{5}^{-\frac{1}{2}}×{2}^{\frac{5}{2}}\\ ={2}^{\frac{1}{2}+\frac{5}{2}}×{5}^{\frac{1}{2}+\left(-\frac{1}{2}\right)}\\ ={2}^{3}+{5}^{\frac{1}{2}-\frac{1}{2}}\\ ={2}^{3}+{5}^{0}\\ =8+1\\ =9\end{array}$  
 
 
 
 
 
Question 3:
 
(a) Find the value of   ${10}^{\frac{4}{3}}÷{10}^{\frac{1}{3}}.$  
 
(b)   Simplify (xy3)5 × x4.
 
 

 Solution:
 
(a)
 
 
  $\begin{array}{l}{10}^{\frac{4}{3}}÷{10}^{\frac{1}{3}}\\ ={10}^{\frac{4}{3}-\frac{1}{3}}\\ ={10}^{\frac{3}{3}}\\ =10\end{array}$  
 

     $\begin{array}{l}\text{(b)}{\left(x{y}^{3}\right)}^{5}×{x}^{4}={x}^{5}{y}^{15}×{x}^{4}\\ \text{}={x}^{5+4}{y}^{15}\\ \text{}={x}^{9}{y}^{15}\end{array}$  
 
 
 
 
 
 
Question 4:
 
(a)   ${\left(81{a}^{8}\right)}^{-\frac{1}{4}}=$  
 
(b)   Find the value of 23 × 22
 
 

 Solution:
 
(a)
 
 
  ${\left(81{a}^{8}\right)}^{-\frac{1}{4}}=\frac{1}{{\left(81{a}^{8}\right)}^{\frac{1}{4}}}=\frac{1}{{\left({3}^{4}\right)}^{\frac{1}{4}}{\left({a}^{8}\right)}^{\frac{1}{4}}}=\frac{1}{3{a}^{2}}$  
 

 (b)
 23 × 22 = 23+2 = 25 = 32
 
 
 
 
 
Question 5:
 
Find the value of the following.
 
(a)   ${81}^{\frac{3}{4}}×{27}^{-1}$  
 
(b)   ${8}^{\frac{2}{3}}×{3}^{-2}$  
 
 

 Solution:
 
(a)
 
  $\begin{array}{l}{81}^{\frac{3}{4}}×{27}^{-1}={\left(\sqrt[4]{81}\right)}^{3}×{\left({3}^{3}\right)}^{-1}\\ \text{}={\left(3\right)}^{3}×{3}^{-3}\\ \text{}={3}^{3+\left(-3\right)}\\ \text{}={3}^{0}=1\end{array}$  
 

 (b)
    $\begin{array}{l}{8}^{\frac{2}{3}}×{3}^{-2}={\left(\sqrt[3]{8}\right)}^{2}×\frac{1}{{3}^{2}}\\ \text{}={\left(2\right)}^{2}×\frac{1}{{3}^{2}}\\ \text{}=4×\frac{1}{9}\\ \text{}=\frac{4}{9}\end{array}$