# 5.2.3 Indices, PT3 Practice

Question 13:

Solution:

Question 14:

Solution:

Question 15:

Solution:

Question 16:

Solution:
$\begin{array}{l}{\left(\frac{-2{x}^{5}{y}^{-2}}{{z}^{\frac{1}{6}}}\right)}^{3}÷\frac{1}{\sqrt{{x}^{2}z}}\\ =\frac{-8{x}^{15}{y}^{-6}}{{z}^{\frac{1}{2}}}×\sqrt{{x}^{2}z}\\ =\frac{-8{x}^{15}{y}^{-6}}{{z}^{\frac{1}{2}}}×{\left({x}^{2}z\right)}^{\frac{1}{2}}\\ =\frac{-8{x}^{15}{y}^{-6}}{\overline{){z}^{\frac{1}{2}}}}×x\overline{){z}^{\frac{1}{2}}}\\ =-8{x}^{15+1}{y}^{-6}\\ =\frac{-8{x}^{16}}{{y}^{6}}\end{array}$

 
 
Question 17:
 
Find the value of the following.
 
(a)   (23)2 × 24 ÷ 25
 
(b)   $\frac{{a}^{2}×{a}^{\frac{1}{2}}}{{\left({a}^{\frac{2}{3}}×{a}^{\frac{1}{3}}\right)}^{-2}}$  
 
 

 Solution:
 
(a)
 
(23)2 × 24 ÷ 25= 26 × 24 ÷ 25
 
= 26+4-5
 
= 25
 
= 32
 
 
 
(b)
 
  $\begin{array}{l}\frac{{a}^{2}×{a}^{\frac{1}{2}}}{{\left({a}^{\frac{2}{3}}×{a}^{\frac{1}{3}}\right)}^{-2}}=\frac{{a}^{2+\frac{1}{2}}}{{\left({a}^{\frac{2}{3}}×{a}^{\frac{1}{3}}\right)}^{-2}}\\ \text{}=\frac{{a}^{2+\frac{1}{2}}}{{\left({a}^{\frac{2}{3}+\frac{1}{3}}\right)}^{-2}}\\ \text{}=\frac{{a}^{\frac{5}{2}}}{{a}^{-2}}\\ \text{}={a}^{\frac{5}{2}-\left(-2\right)}\\ \text{}={a}^{\frac{5}{2}+\frac{4}{2}}\\ \text{}={a}^{\frac{9}{2}}\end{array}$  
 
 
 

# 5.2.2 Indices, PT3 Practice

Question 7:
Given that ${2}^{8-x}\text{=32}$ , calculate the value of x.

Solution:
$\begin{array}{l}{2}^{8-x}\text{=32}\\ {2}^{8-x}{\text{=2}}^{5}\\ 8-x=5\\ -x=-3\\ x=3\end{array}$

Question 8:

Solution:
$\begin{array}{l}{3}^{2p-1}\text{=}\left({3}^{p}\right)\left({3}^{2}\right)\\ {3}^{2p-1}\text{=}{3}^{p+2}\\ 2p-1=p+2\\ p=3\end{array}$

Question 9:

Solution:
$\begin{array}{l}\text{8}×{\text{8}}^{p+1}\text{=}\left({8}^{5}\right)\left({8}^{3}\right)\\ {8}^{1+p+1}={8}^{5+3}\\ 2+p=8\\ p=6\end{array}$

Question 10:

Solution:
$\begin{array}{l}\frac{{\text{2}}^{5}×{2}^{7}}{{2}^{10}}={2}^{p}\\ {\text{2}}^{5+7-10}={2}^{p}\\ {\text{2}}^{2}={2}^{p}\\ p=2\end{array}$

Question 11:

Solution:

Question 12:

Solution:

# 5.2.1 Indices, PT3 Practice

 
 
5.2.1 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)
    $\begin{array}{l}{8}^{\frac{2}{3}}×{3}^{-2}={\left(\sqrt[3]{8}\right)}^{2}×\frac{1}{{3}^{2}}\\ \text{}={\left(2\right)}^{2}×\frac{1}{{3}^{2}}\\ \text{}=4×\frac{1}{9}\\ \text{}=\frac{4}{9}\end{array}$  
 
 
 
 
 
Question 6:
 
Find the value of the following.
 
(a)   ${8}^{\frac{4}{3}}×{\left({3}^{-2}\right)}^{3}×{9}^{\frac{3}{2}}$  
 
(b)   $\frac{{2}^{-2}×{3}^{2}}{{2}^{-3}×81}$  
 
 

 Solution:
 
(a)
 
 
  $\begin{array}{l}{8}^{\frac{4}{3}}×{\left({3}^{-2}\right)}^{3}×{9}^{\frac{3}{2}}\\ ={\left({2}^{3}\right)}^{\frac{4}{3}}×{3}^{-6}×{\left({3}^{2}\right)}^{\frac{3}{2}}\\ ={2}^{4}×{3}^{-6}×{3}^{3}\\ =16×{3}^{-6+3}\\ =16×{3}^{-3}\\ =16×\frac{1}{{3}^{3}}\\ =\frac{16}{27}\end{array}$  
 
 
(b)
 
  $\begin{array}{l}\frac{{2}^{-2}×{3}^{2}}{{2}^{-3}×81}=\frac{{2}^{-2}×{3}^{2}}{{2}^{-3}×{3}^{4}}\\ \text{}={2}^{-2-\left(-3\right)}×{3}^{2-4}\\ \text{}=2×{3}^{-2}\\ \text{}=\frac{2}{{3}^{2}}\\ \text{}=\frac{2}{9}\end{array}$  
 
 
 
 

# 5.1 Indices

5.1 Indices

5.1.1 Indices
1.   A number expressed in the form an is known as an index notation.
2.   an is read as ‘a to the power of n’ where a is the base and n is the index.
Example:

3.
If a is a real number and is a positive integer, then

4.   The value of a real number in index notation can be found by repeated multiplication.
Example:
6= 6 × 6 × 6 × 6
= 1296

5.
A number can be expressed in index notation by dividing the number repeatedly by the base.
Example:
243 = 3 × 3 × 3 × 3 × 3
= 35

5.1.2 Multiplication of Numbers in Index Notation
The multiplication of numbers or algebraic terms with the same base can be done by using the Law of Indices.

am  × an = am + n

Example:
33 × 38 = 33+8
= 311

5.1.3 Division of Numbers in Index Notation
1. The Law of indices for division is:

am  ÷ an = am - n

Example:
412 ÷ 412 = 412-12
= 40
= 1

2. a0 = 1

5.1.4 Raise Numbers and Algebraic Terms in Index Notation to a Power
To raise a number in index notation to a power, multiply the two indices while keeping the base unchanged.

(am ) n  = (an ) m  = amn

Example:
(43)7 = 43×7
= 421

5.1.5 Negative Indices

${a}^{-n}=\frac{1}{{a}^{n}}$

Example:
${5}^{-2}=\frac{1}{{5}^{2}}=\frac{1}{25}$

5.1.6 Fractional Indices

Example: