5.2.3 Indices, PT3 Practice

Question 13:
$\text{Solve: }{3}^{2n+1}=\frac{3^n\times 9}{{\left(9^{\frac{1}{2}}\right)}^{-3}}$

Solution:
$\begin{array}{l}{3}^{2n+1}=\frac{3^n\times 9}{{\left(9^{\frac{1}{2}}\right)}^{-3}}\\ 3^{2n+1}=\frac{3^n\times 3^2}{3^{-3}}\\ 3^{2n+1}=3^{n+2-\left(-3\right)}\\ 2n+1=n+5\\ n=4\end{array}$

Question 14:
$\text{Given that }{k}^3=9^{-\frac{3}{2}}\times {64}^{\frac{1}{2}},\text{ find the value of }k.$

Solution:
$\begin{array}{l}{k}^3=9^{-\frac{3}{2}}\times {64}^{\frac{1}{2}}\\ ={\left(3^2\right)}^{-\frac{3}{2}}\times {\left(2^6\right)}^{\frac{1}{2}}\\ =3^{-3}\times 2^3\\ ={\left(\frac{2}{3}\right)}^3\\ k=\frac{2}{3}\end{array}$

Question 15:
${\text{Given 9}}^{x+2}÷3^4=3^{x+1},\text{ calculate the value of }x.$

Solution:
$\begin{array}{l}{\text{9}}^{x+2}÷3^4=3^{x+1}\\ {\left(3^2\right)}^{x+2}÷3^4=3^{x+1}\\ 3^{2x+4}÷3^4=3^{x+1}\\ 2x+4-4=x+1\\ 2x=x+1\\ x=1\end{array}$

Question 16:
$\text{Simplify: }{\left(\frac{-2x^5{y}^{-2}}{z^{\frac{1}{6}}}\right)}^3÷\frac{1}{\sqrt{x^{2}z}}$

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