1. ## 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

2. 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:

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

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

3. How would you finish off what you've started?

4. Originally Posted by freswood
How would you finish off what you've started?
$\displaystyle 6=e \times e^{b}+B$
$\displaystyle 2=e^b+B$

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

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

6. Hello, freswood!

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

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

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