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: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
$\displaystyleOriginally Posted by freswood
6=e \times e^{b}+B$
$\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
Hello, freswood!
This is a variation of Captain Black's solution . . .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: .$\displaystyle b\:=\:\ln\left(\frac{4}{e-1}\right)$