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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- Jun 7th 2006, 09:49 PMfreswoodsimultaneous 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 - Jun 7th 2006, 10:45 PMCaptainBlackQuote:

Originally Posted by**freswood**

$\displaystyle

6=e^{1+b}+B$

$\displaystyle

2=e^b+B

$

you rewrite them as:

$\displaystyle

6=e \times e^{b}+B$

$\displaystyle

2=e^b+B

$,

which leaves you with a pair of linear simultaneous equations in $\displaystyle e^b$ and $\displaystyle B$,

which can be solved in the usual manner.

RonL - Jun 7th 2006, 11:02 PMfreswood
How would you finish off what you've started?

- Jun 7th 2006, 11:10 PMCaptainBlackQuote:

Originally Posted by**freswood**

6=e \times e^{b}+B$

$\displaystyle

2=e^b+B

$

Subtract:

$\displaystyle

6-2=(e-1) \times e^{b}+B-B$,

so:

$\displaystyle

e^b=\frac{4}{e-1}

$,

and substituting back into the second equation:

$\displaystyle

B=2-\frac{4}{e-1}

$.

Finally:

$\displaystyle

b=\log_e(4/(e-1))

$

RonL - Jun 8th 2006, 01:00 AMfreswood
Thanks so much! It kinda confused me because I wasn't expecting the answer to be that messy. Whole numbers all the way! :D

- Jun 8th 2006, 06:39 PMSoroban
Hello, freswood!

Quote:

Solve: .$\displaystyle \begin{array}{cc}(1)\;e^{b+1} + B \:= \:6 \\ (2) \;e^b +B \:= \:2\end{array}$

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

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

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

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

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

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

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

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