Few more Algebra problems..

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June 19th 2006, 08:34 PM
daniel4616

Few more Algebra problems..

I have a couples problems with certain types of problems. I cant seem how to figure these ones out, any help is appreciated.

1. Find x and express answer in terms of natural logarithms: 2e^3x=6

2. Express 1+ ln as a single logarithm.

3. Write log(x+3) in terms of natural logarithms.

4. Solve with Gauss Jordan method (I can get half way done but I never seem to be able to finish the whole problem.)
x-y-3z=2
2x-y-4z=3
x+y-z=1

5. How to Maximize Z=2x-3y subject to
2x+y>=1
x-y=<1
x,y>=0

6. Similar can probably figure it with some help from previous one.
Maximize Z=2x-3y subject to
-x+y=<2
3x+y=<18
x,y>=0

All help appreciated.
June 19th 2006, 08:47 PM
malaygoel

Quote:

Originally Posted by daniel4616

1. Find x and express answer in terms of natural logarithms: 2e^3x=6

$2e^{3x} =6$
We can divide both sides by 2(not equal to zero)
$e^{3x} =3$
Taking natural log both sides
$3x =\ln3$
$x =\frac{\ln3}{3}$
$x =\frac{1}{3}*\ln3$
$x =\ln{3^{1/3}}$
$x =\ln{\sqrt[3]{3}}$
Keep Smiling
Malay
June 19th 2006, 08:51 PM
malaygoel

Quote:

Originally Posted by daniel4616

2. Express 1+ ln as a single logarithm.

$\ln$ does not mean anything alone, please complete the question
I assume it is $\ln{x}(x>0)$
$1 + \ln{x} = \ln{e} + \ln{x}= \ln{ex}$
June 20th 2006, 04:49 AM
Soroban

Hello, daniel4616!

Here's #4 . . .

Quote:

4. Solve with Gauss-Jordan method: . $\begin{array}{ccc}x-y-3z\,=\,2\\2x-y-4z\,=\,3\\x+y-z\,=\,1\end{array}$

We have: . $\begin{pmatrix}1 & \text{-}1 & \text{-}3 & | & 2 \\ 2 & \text{-}1 & \text{-}4 & | & 3 \\ 1 & 1 & \text{-}1 & | & 1\end{pmatrix}$

$\begin{array}{ccc} \\ R_2\,\text{-}\,2\!\cdot\!R_1\\ R_3\,\text{-}\,R_1\end{array}\;\begin{pmatrix}1 & \text{-}1 & \text{-}3 & | & 2\\0 & 1 & 2 & | & \text{-}1\\0 & 2 & 2 & | & \text{-}1\end{pmatrix}$

$\begin{array}{ccc}R_1\,\text{+}\,R_2\\ \\ R_3\,\text{-}\,2\!\cdot\!R_2\end{array}\;\begin{pmatrix}1 & 0 & \text{-}1 & | & 1 \\ 0 & 1 & 2 & | & \text{-}1\\ 0 & 0 & \text{-}2 & | & 1\end{pmatrix}$

$\begin{array}{ccc} \\ \\ R_3\div(-2)\end{array}\;\begin{pmatrix}1 & 0 & \text{-}1 & | & 1 \\ 0 & 1 & 2 & | & \text{-}1 \\ 0 & 0 & 1 & | & \text{-}\frac{1}{2}\end{pmatrix}$

$\begin{array}{ccc}R_1\,\text{+}\,R_3\\ R_2\,\text{-}\,2\!\cdot\!R_3 \\ \end{array}\;\begin{pmatrix}1 & 0 & 0 & | & \frac{1}{2} \\0 & 1 & 0 & | & 0 \\ 0 & 0 & 1 & | & \text{-}\frac{1}{2}\end{pmatrix}$
June 20th 2006, 05:54 AM
Soroban

1 Attachment(s)

Hello again, daniel4616!

Here's #6 . . . in baby-steps . . .

Quote:

6. Maximize $z\:=\:2x-3y$ subject to: $\begin{Bmatrix}[1]\;\;\quad x,y \geq 0\quad\;\; \\ [2]\;\; -x + y\:\leq \:2\\ [3]\;\; 3x + y \:\leq\:18\end{Bmatrix}$

We must graph and shade the inequalities.

$[1]\;x,y \geq 0$ places us in the first quadrant.

$[2]\;-x + y \:\leq \:2$

Graph the line: $-x + y\:=\:2$
It has intercepts: $(-2,0),\;(0,2)$ . . Sketch the line.

Solve for $y$ and note the inequality: . $y \:\leq\:x + 2$
Since the inequality is $\leq$ , shade the region below the line.

Code:

| *: | *:::: | *::::::: 2*:::::::::: * |::::::::::: - - * - + - - - - - - - - -2 |

$[3]\;3x + y \:\leq \:18$
Graph the line $3x + y \:=\:18$
It has intercepts: $(6,0),\;(0,18)$

Since we have: $y \:\leq\:-3x + 18$ , we shade below the line.

Code:

| 18* |* |:* |::* |:::* |::::* |:::::* |::::::* - + - - - *- - | 6

Of course, we graph these region on the same graph.
And the final shaded region looks like this.
(Note the vertices of the region.)

Code:

| | *(4,6) | *::* (0,2)*:::::* |:::::::* - - * - - - - * - - (0,0) (6.0)

We find the intersection of the two slanted lines
. . by solving the system: $\begin{array}{cc}-x + y\:=\:2 \\ 3x + y \:= \:18\end{array}$
and we get: $x = 4,\;y = 6$

The vertices are the only "critical values" we are concerned with.
Test them in the z-function to see which gives a maximum value.

$(0,0):\;z\:=\:2\!\cdot\!0 - 3\!\cdot\!0\:=\:0$
$(6,0):\;z\:=\:2\!\cdot\!6 - 3\!\cdot\!0\:=\:12\quad\Leftarrow\;maximum!$
$(0,2):\;z\:=\:2\!\cdot\!0 - 3\!\cdot\!2\:=\:-6$
$(4,6):\;z\:=\:2\!\cdot\!4 - 3\!\cdot\!6\:=\:-10$

Answer: . $x = 6,\;y = 0$ to maximize $z$ .

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