Simultaneous log equations

• March 27th 2010, 07:30 AM
Stuck Man
Simultaneous 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.
• March 27th 2010, 07:36 AM
e^(i*pi)
Quote:

Originally Posted by Stuck Man
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 $xy^2 = 3$

$x = \frac{3}{y^2}$

$[log_3 \left(\frac{3}{y^2}\right)][\log_3(y)] = (1-2\log_3(y))(\log_3(y)) = -3$

This simplifies to $2[\log_3(y)]^2 - \log_3(y)-3=0$ which is a quadratic equation, hence use the formula to solve for $\log_3(y)$ given that it must be greater than 0
• March 27th 2010, 07:54 AM
Stuck 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.
• March 27th 2010, 07:58 AM
Stuck Man
The correct scan is attached to this.
• March 27th 2010, 08:18 AM
Stuck Man
I'm not sure if the final line is any use on the left, x=-2y+3.
• March 27th 2010, 01:49 PM
Soroban
Helolo, Stuck Man!

Quote:

17. Show that: . $\log_9(xy^2) \:=\:\tfrac{1}{2}\log_3x + \log_3y$

Hence, solve: . $\begin{array}{ccc}\log_9(xy^2) &=& \frac{1}{2} \\ \\[-3mm] (\log_3x)(\log_3y) &=& \text{-}3 \end{array}$

Base-change formula:

. . $\log_9(xy^2) \;=\;\frac{\log_3(xy^2)}{\log_39} \;=\;\frac{\log_3x+ \log_3y^2}{2} \;=\;\frac{\log_3x + 2\log_3y}{2} \;=\;\tfrac{1}{2}\log_3x + \log_3y$

The system becomes: . $\begin{array}{ccc} \frac{1}{2}\log_3x + \log_3y &=& \frac{1}{2} \\ \\[-3mm] (\log_3x)(\log_3y) &=& \text{-}3 \end{array}$

Let: . $\begin{Bmatrix} X &=& \log_3x \\ Y &=& \log_3y \end{Bmatrix} \quad\hdots\quad \text{And we have: }\;
\begin{Bmatrix}\frac{1}{2}X + Y &=& \frac{1}{2} \\ \\[-3mm] XY &=& \text{-}3 \end{Bmatrix}$
. . $\Rightarrow\quad \begin{Bmatrix}X + 2Y &=& 1 & [1] \\ XY &=& \text{-}3 & [2] \end{Bmatrix}$

From [1], we have: . $X \:=\:1 - 2Y$

Substitute into [2]: . $(1-2Y)Y \:=\:-3 \quad\Rightarrow\quad 2Y^2 - Y - 3 \:=\:0 \quad\Rightarrow\quad (Y+1)(2Y-3) \:=\:0$

Hence: . $Y \:=\:-1,\:\tfrac{3}{2}\quad\Rightarrow\quad X \:=\:3,\:-2$

Back-substitute:

$(X,Y) \:=\:(3,\text{-}1)\!:\;\;\begin{array}{cccccccccc} \log_3x \:=\: 3 & \Rightarrow & x \:=\: 3^3 & \Rightarrow & x \:=\:27 \\ \log_3y \:=\:\text{-}1 & \Rightarrow &y \:=\:3^{-1} & \Rightarrow & y \:=\:\frac{1}{3} \end{array}$

$(X,Y) \:=\:\left(\text{-}2,\tfrac{3}{2}\right)\!:\;\;\begin{array}{cccccc} \log_3x \:=\:\text{-}2 & \Rightarrow & x \:=\:3^{-2} & \Rightarrow & x \:=\:\frac{1}{9} \\ \log_3y \:=\:\frac{3}{2} & \Rightarrow & y \:=\:3^{\frac{3}{2}} & \Rightarrow & y \:=\:3\sqrt{3} \end{array}$

Therefore: . $(x,y) \;=\;\left(27,\:\frac{1}{3}\right),\;\left(\frac{1 }{9},\:3\sqrt{3}\right)$

• March 28th 2010, 07:52 AM
Stuck Man
Thanks. I thought there might be different methods of doing the question.