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Math Help - Few more Algebra problems..

  1. #1
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    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.
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  2. #2
    Super Member malaygoel's Avatar
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    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
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  3. #3
    Super Member malaygoel's Avatar
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    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}
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  4. #4
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    Hello, daniel4616!

    Here's #4 . . .


    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}

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  5. #5
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    Hello again, daniel4616!

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


    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.


    -----
    Attached Thumbnails Attached Thumbnails Few more Algebra problems..-picture10.gif  
    Last edited by ThePerfectHacker; June 20th 2006 at 08:41 AM.
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