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.

Printable View

- Mar 27th 2010, 06:30 AMStuck ManSimultaneous log equations
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.

- Mar 27th 2010, 06:36 AMe^(i*pi)
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 - Mar 27th 2010, 06:54 AMStuck Man
I wasn't expecting the way you started it. I did something more like the earlier part of the question. I will scan my work in a minute.

- Mar 27th 2010, 06:58 AMStuck Man
The correct scan is attached to this.

- Mar 27th 2010, 07:18 AMStuck Man
I'm not sure if the final line is any use on the left, x=-2y+3.

- Mar 27th 2010, 12:49 PMSoroban
Helolo, Stuck Man!

Quote:

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) $

- Mar 28th 2010, 06:52 AMStuck Man
Thanks. I thought there might be different methods of doing the question.