**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:**$\text{Given}{3}^{2p-1}\text{=}\left({3}^{p}\right)\left({3}^{2}\right)\text{,calculatethevalueof}p.$

*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:**$\text{Giventhat8}\times {\text{8}}^{p+1}\text{=}\left({8}^{5}\right)\left({8}^{3}\right)\text{,findthevalueof}p.$

*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:**$\text{Giventhat}\frac{{\text{2}}^{5}\times {2}^{7}}{{2}^{10}}={2}^{p}\text{,findthevalueof}p.$

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

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

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

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