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Math Help - simultaneous equations

  1. #1
    Junior Member
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    Smile simultaneous equations

    How would you go about solving:

    6 = e^(1+b) + B
    2 = e^b +B

    where e = euler's number, not an unknown

    Is it something to do with subtracting? I just keep getting myself into a dead end.

    Thanks
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by freswood
    How would you go about solving:

    6 = e^(1+b) + B
    2 = e^b +B

    where e = euler's number, not an unknown

    Is it something to do with subtracting? I just keep getting myself into a dead end.

    Thanks
    To solve:

    <br />
6=e^{1+b}+B
    <br />
2=e^b+B<br />

    you rewrite them as:

    <br />
6=e \times e^{b}+B
    <br />
2=e^b+B<br />
,

    which leaves you with a pair of linear simultaneous equations in e^b and B,
    which can be solved in the usual manner.

    RonL
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  3. #3
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    How would you finish off what you've started?
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  4. #4
    Grand Panjandrum
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    Quote Originally Posted by freswood
    How would you finish off what you've started?
    <br />
6=e \times e^{b}+B
    <br />
2=e^b+B<br />

    Subtract:

    <br />
6-2=(e-1) \times e^{b}+B-B,

    so:

    <br />
e^b=\frac{4}{e-1}<br />
,

    and substituting back into the second equation:

    <br />
B=2-\frac{4}{e-1}<br />
.

    Finally:

    <br />
b=\log_e(4/(e-1))<br />

    RonL
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  5. #5
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    Thanks so much! It kinda confused me because I wasn't expecting the answer to be that messy. Whole numbers all the way!
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  6. #6
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    Hello, freswood!

    Solve: . \begin{array}{cc}(1)\;e^{b+1} + B \:= \:6 \\  (2) \;e^b +B \:= \:2\end{array}
    This is a variation of Captain Black's solution . . .

    We have: . \begin{array}{cc}e^{b+1}\:=\:6 - B \\ e^b\:=\:2 - B\end{array}

    Divide the equations: . \frac{e^{b+1}}{e^b}\;=\;\frac{6 - B}{2 - B}\quad\Rightarrow\quad e\;=\;\frac{6 - B}{2 - B}

    . . Then: . e(2 - B)\:=\:6 - B\quad\Rightarrow\quad 2e - Be\:=\:6 - B

    . . And: . 2e - 6 \:=\:Be - B\quad\Rightarrow\quad 2(e - 3)\:=\:B(e - 1)

    . . Therefore: . B\:=\:\frac{2(e-3)}{e-1}


    Substitute into (2): . e^b - \frac{2(e-3)}{e - 1}\:=\:2\quad\Rightarrow\quad e^b \:=\:2 - \frac{2(e - 3)}{e - 1}

    . . . e^b\:=\:\frac{2(e - 1) - 2(e - 3)}{e - 1} \quad\Rightarrow\quad e^b \:=\:\frac{4}{e - 1}

    . . Therefore: . b\:=\:\ln\left(\frac{4}{e-1}\right)
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