Question 9:
$$ (a) \text { Given that } 2^x=8^y=\sqrt{2^{3 z}} \text {, find } x: y: z \text {. }{[2 marks]} $$
$$ (b) \text { Given that } 9^{2pq+1}+3^{4 pq-3}=732 \text {, express } p \text { in terms of } q \text {. }{[3 marks]} $$
$$ (c) \text { Find the value of } \frac{1}{\log _a a b}+\frac{1}{\log _b a b} \text {. }{[2 marks]} $$
Solution:
(a)
$$ \begin{aligned} 2 x^2 & =8 y=\sqrt{2^{3 z}} \\ 2^x & =\left(2^3\right) y=\left(2^{3 z}\right)^{\frac{1}{2}} \\ 2^x & =2^{3 y}=2^{\frac{3 x}{2}} \\ x & =3 y=\frac{3 z}{2} \end{aligned} $$
$$ \begin{aligned} & x=3 y \\ & \frac{x}{y}=\frac{3}{1} \\ & x: y=3: 1 \end{aligned} $$
$$ \begin{aligned} & 3 y=\frac{3 z}{2} \\ & \frac{y}{z}=\frac{\left(\frac{3}{2}\right)}{3} \\ & \frac{y}{z}=\frac{1}{2} \\ & y: z=1: 2 \end{aligned} $$
$$ \begin{aligned} & x: y: z \\ & 3: 1 \\ & \quad 1: 2 \\ & \therefore x: y: z=3: 1: 2 \end{aligned} $$
(b)
$$ \begin{aligned} 9^{2 p q+1}+3^{4 p q-3} & =732 \\ 3^{2(2 p q+1)}+3^{4 p q-3} & =732 \\ 3^{4 p q+2}+3^{4 p q-3} & =732 \\ \left(3^{4 p q}\right)\left(3^2\right)+\left(3^{4 p q}\right)\left(3^{-3}\right) & =732 \\ 3^{4 p q}\left(3^2+3^{-3}\right) & =732 \\ 3^{4 p q}\left(9+\frac{1}{27}\right) & =732 \\ 3^{4 p q}\left(\frac{244}{27}\right) & =732 \\ 3^{4 p q} & =732 \div \frac{244}{27} \\ 3^{4 p q} & =81 \\ 3^{4 p q} & =3^4 \\ 4 p q & =4 \\ p q & =1 \\ p & =\frac{1}{q} \end{aligned} $$
(c)
$$ \begin{aligned} \frac{1}{\log _a a b}+\frac{1}{\log _b a b} & =\log _{a b} a+\log _{a b} b \\ & =\log _{a b}(a \times b) \\ & =\log _{a b} a b \\ & =1 \end{aligned} $$
$$ (a) \text { Given that } 2^x=8^y=\sqrt{2^{3 z}} \text {, find } x: y: z \text {. }{[2 marks]} $$
$$ (b) \text { Given that } 9^{2pq+1}+3^{4 pq-3}=732 \text {, express } p \text { in terms of } q \text {. }{[3 marks]} $$
$$ (c) \text { Find the value of } \frac{1}{\log _a a b}+\frac{1}{\log _b a b} \text {. }{[2 marks]} $$
Solution:
(a)
$$ \begin{aligned} 2 x^2 & =8 y=\sqrt{2^{3 z}} \\ 2^x & =\left(2^3\right) y=\left(2^{3 z}\right)^{\frac{1}{2}} \\ 2^x & =2^{3 y}=2^{\frac{3 x}{2}} \\ x & =3 y=\frac{3 z}{2} \end{aligned} $$
$$ \begin{aligned} & x=3 y \\ & \frac{x}{y}=\frac{3}{1} \\ & x: y=3: 1 \end{aligned} $$
$$ \begin{aligned} & 3 y=\frac{3 z}{2} \\ & \frac{y}{z}=\frac{\left(\frac{3}{2}\right)}{3} \\ & \frac{y}{z}=\frac{1}{2} \\ & y: z=1: 2 \end{aligned} $$
$$ \begin{aligned} & x: y: z \\ & 3: 1 \\ & \quad 1: 2 \\ & \therefore x: y: z=3: 1: 2 \end{aligned} $$
(b)
$$ \begin{aligned} 9^{2 p q+1}+3^{4 p q-3} & =732 \\ 3^{2(2 p q+1)}+3^{4 p q-3} & =732 \\ 3^{4 p q+2}+3^{4 p q-3} & =732 \\ \left(3^{4 p q}\right)\left(3^2\right)+\left(3^{4 p q}\right)\left(3^{-3}\right) & =732 \\ 3^{4 p q}\left(3^2+3^{-3}\right) & =732 \\ 3^{4 p q}\left(9+\frac{1}{27}\right) & =732 \\ 3^{4 p q}\left(\frac{244}{27}\right) & =732 \\ 3^{4 p q} & =732 \div \frac{244}{27} \\ 3^{4 p q} & =81 \\ 3^{4 p q} & =3^4 \\ 4 p q & =4 \\ p q & =1 \\ p & =\frac{1}{q} \end{aligned} $$
(c)
$$ \begin{aligned} \frac{1}{\log _a a b}+\frac{1}{\log _b a b} & =\log _{a b} a+\log _{a b} b \\ & =\log _{a b}(a \times b) \\ & =\log _{a b} a b \\ & =1 \end{aligned} $$