1.2.2 Indices, PT3 Practice

 
 
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)
 
  $\begin{array}{l}\frac{{2}^{-2}×{3}^{2}}{{2}^{-3}×81}=\frac{{2}^{-2}×{3}^{2}}{{2}^{-3}×{3}^{4}}\\ \text{}={2}^{-2-\left(-3\right)}×{3}^{2-4}\\ \text{}=2×{3}^{-2}\\ \text{}=\frac{2}{{3}^{2}}\\ \text{}=\frac{2}{9}\end{array}$  
 
 
 
 
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:
$\begin{array}{l}\text{8}×{\text{8}}^{p+1}\text{=}\left({8}^{5}\right)\left({8}^{3}\right)\\ {8}^{1+p+1}={8}^{5+3}\\ 2+p=8\\ p=6\end{array}$