simultaneous equations

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June 7th 2006, 09:49 PM
freswood

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
June 7th 2006, 10:45 PM
CaptainBlack

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:

$ 6=e^{1+b}+B$
$ 2=e^b+B $

you rewrite them as:

$ 6=e \times e^{b}+B$
$ 2=e^b+B $ ,

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

RonL
June 7th 2006, 11:02 PM
freswood

How would you finish off what you've started?
June 7th 2006, 11:10 PM
CaptainBlack

Quote:

Originally Posted by freswood

How would you finish off what you've started?

$ 6=e \times e^{b}+B$
$ 2=e^b+B $

Subtract:

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

so:

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

and substituting back into the second equation:

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

Finally:

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

RonL
June 8th 2006, 01:00 AM
freswood

Thanks so much! It kinda confused me because I wasn't expecting the answer to be that messy. Whole numbers all the way! :D
June 8th 2006, 06:39 PM
Soroban

Hello, freswood!

Quote:

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