# 1.2.4 Indices, PT3 Practice

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