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:
$\text{Given }{3}^{2p-1}\text{=}\left(3^p\right)\left(3^2\right)\text{, calculate the value of }p.$

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:
$\text{Given that 8}\times {\text{8}}^{p+1}\text{=}\left(8^5\right)\left(8^3\right)\text{, find the value of }p.$

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

Question 10:
$\text{Given that }\frac{{\text{2}}^5\times 2^7}{2^{10}}=2^p\text{, find the value of }p.$

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

Question 11:
$\text{Simplify: }\frac{12p^{10}{q}^6}{3p^4{q}^2\times {\left(4p{q}^3\right)}^2}$

Solution:
$\begin{array}{l}\frac{12p^{10}{q}^6}{3p^4{q}^2\times {\left(4p{q}^3\right)}^2}=\frac{12p^{10}{q}^6}{3p^4{q}^2\times 16p^2{q}^6}\\ =\frac{{\cancel{12}}^1{p}^{10-4-2}{q}^{6-2-6}}{{\cancel{3}}^1\times {\cancel{16}}^4}\\ =\frac{p^4{q}^{-2}}{4}\\ =\frac{p^4}{4q^2}\end{array}$

Question 12:
$\text{Simplify: }\frac{a{b}^2\times {\left(8a^3{b}^6\right)}^{\frac{1}{3}}}{{\left(a^2{b}^4\right)}^{\frac{1}{2}}}$

Solution:
$\begin{array}{l}\frac{a{b}^2\times {\left(8a^3{b}^6\right)}^{\frac{1}{3}}}{{\left(a^2{b}^4\right)}^{\frac{1}{2}}}=\frac{a{b}^2\times \left(8^{{}^{\frac{1}{3}}}{a}^3{}^{\left(\frac{1}{3}\right)}{b}^{6\left(\frac{1}{3}\right)}\right)}{a^{2\left(\frac{1}{2}\right)}{b}^{4\left(\frac{1}{2}\right)}}\\ =\frac{\cancel{a{b}^2}\times 2a{b}^2}{\cancel{a{b}^2}}\\ =2a{b}^2\end{array}$