I have solved the simultaneous equations but I think there might have been an easier method. I have x=-2y+3 from the first equation. I don't think its correct. I substituted part of the first equation into the second one.
What was the first part of the question? The question is suggesting you use the answer to the previous part of the question to solve this one.
We have $\displaystyle xy^2 = 3$
$\displaystyle x = \frac{3}{y^2}$
$\displaystyle [log_3 \left(\frac{3}{y^2}\right)][\log_3(y)] = (1-2\log_3(y))(\log_3(y)) = -3$
This simplifies to $\displaystyle 2[\log_3(y)]^2 - \log_3(y)-3=0$ which is a quadratic equation, hence use the formula to solve for $\displaystyle \log_3(y)$ given that it must be greater than 0
Helolo, Stuck Man!
17. Show that: . $\displaystyle \log_9(xy^2) \:=\:\tfrac{1}{2}\log_3x + \log_3y$
Hence, solve: .$\displaystyle \begin{array}{ccc}\log_9(xy^2) &=& \frac{1}{2} \\ \\[-3mm] (\log_3x)(\log_3y) &=& \text{-}3 \end{array}$
Base-change formula:
. . $\displaystyle \log_9(xy^2) \;=\;\frac{\log_3(xy^2)}{\log_39} \;=\;\frac{\log_3x+ \log_3y^2}{2} \;=\;\frac{\log_3x + 2\log_3y}{2} \;=\;\tfrac{1}{2}\log_3x + \log_3y$
The system becomes: .$\displaystyle \begin{array}{ccc} \frac{1}{2}\log_3x + \log_3y &=& \frac{1}{2} \\ \\[-3mm] (\log_3x)(\log_3y) &=& \text{-}3 \end{array}$
Let: .$\displaystyle \begin{Bmatrix} X &=& \log_3x \\ Y &=& \log_3y \end{Bmatrix} \quad\hdots\quad \text{And we have: }\;
\begin{Bmatrix}\frac{1}{2}X + Y &=& \frac{1}{2} \\ \\[-3mm] XY &=& \text{-}3 \end{Bmatrix}$ . . $\displaystyle \Rightarrow\quad \begin{Bmatrix}X + 2Y &=& 1 & [1] \\ XY &=& \text{-}3 & [2] \end{Bmatrix}$
From [1], we have: .$\displaystyle X \:=\:1 - 2Y$
Substitute into [2]: .$\displaystyle (1-2Y)Y \:=\:-3 \quad\Rightarrow\quad 2Y^2 - Y - 3 \:=\:0 \quad\Rightarrow\quad (Y+1)(2Y-3) \:=\:0 $
Hence: .$\displaystyle Y \:=\:-1,\:\tfrac{3}{2}\quad\Rightarrow\quad X \:=\:3,\:-2$
Back-substitute:
$\displaystyle (X,Y) \:=\:(3,\text{-}1)\!:\;\;\begin{array}{cccccccccc} \log_3x \:=\: 3 & \Rightarrow & x \:=\: 3^3 & \Rightarrow & x \:=\:27 \\ \log_3y \:=\:\text{-}1 & \Rightarrow &y \:=\:3^{-1} & \Rightarrow & y \:=\:\frac{1}{3} \end{array}$
$\displaystyle (X,Y) \:=\:\left(\text{-}2,\tfrac{3}{2}\right)\!:\;\;\begin{array}{cccccc} \log_3x \:=\:\text{-}2 & \Rightarrow & x \:=\:3^{-2} & \Rightarrow & x \:=\:\frac{1}{9} \\ \log_3y \:=\:\frac{3}{2} & \Rightarrow & y \:=\:3^{\frac{3}{2}} & \Rightarrow & y \:=\:3\sqrt{3} \end{array}$
Therefore: .$\displaystyle (x,y) \;=\;\left(27,\:\frac{1}{3}\right),\;\left(\frac{1 }{9},\:3\sqrt{3}\right) $