Author Topic: soving simulataneous equations for exponential functions? (Read 1121 times) Tweet Share

tolga · « **on:** January 02, 2010, 07:07:14 pm »

Solve simultaneously for b:

2=e^b[1]
6=e^1+b[2]

Sove simultanoesly for A and k:

-10=Ae^-2k[1]
-4=Ae^-k[2]

Solve simultaneously for A and k:

1=Alog10(k)[1] Note: the 10 after log is to the power below log
2=Alog10(3k)[2]

kriptik · « **Reply #1 on:** January 02, 2010, 07:08:31 pm »

use tex, its hurting my eyes..

brightsky · « **Reply #2 on:** January 02, 2010, 07:19:39 pm »

$2 = e^b$

From this, you already know that:

$b = log_e2$

No simultaneous equations needed.

GerrySly · « **Reply #3 on:** January 02, 2010, 07:28:35 pm »

For the second one...

$\begin{align*} -10&=\text{A}\cdot e^{-2k}\\ -4&=\text{A}\cdot e^{-k} \end{align*}$

First one divided by the second leaves...

$\begin{align*} \frac{5}{2}&=e^{-k}\\ k&=\log_e{\left ( \frac{2}{5} \right )} \end{align*}$

Sub that back into the second equation...

$\begin{align*} -4&=\text{A}\cdot e^{-\log_e{\left ( \frac{2}{5} \right )}}\\ &=\frac{5A}{2}\\ A&=-\frac{8}{5} \end{align*}$

I'm sure you can do the last one

brightsky · « **Reply #4 on:** January 02, 2010, 07:32:25 pm »

$-10=Ae^{-2k}$ [1]

$-4=Ae^{-k}$ [2]

From [2],

$\frac{-4}{A} = e^{-k}$

$\left(\frac {-4}{A}\right) = (e^{-k})^2$ ==> Can someone confirm if this is a legit step?

$\frac {16}{A^2} = e^{-2k}$ [3]

From [1],

$\frac {-10}{A} = e^{-2k}$ [4]

From [3] and [4],

$\frac {-10}{A} = {16}{A^2}$

$16A = -10A^2$

$16 = -10A$

$A = \frac {16}{-10} = -\frac {8}{5}$

Substitute this back into [1]:

$-10 = \frac{8}{5} e^{-2k}$

$\frac{25}{4} = e^{-2k}$

$log_e(\frac{25}{4}) = -2k$

$log_e(\frac{5}{2})^2 = -2k$

$2 log_e (\frac{5}{2}) = -2k$

$log_e(5) - log_e(2) = -k$

$k = log_e(2) - log_e(5)$

Edit: GerrySly's way is quicker.

GerrySly · « **Reply #5 on:** January 02, 2010, 07:56:41 pm »

Quote from: brightsky on January 02, 2010, 07:39:17 pm

$1=Alog_{10}(k)$ [1]

$2=Alog_{10}(3k)$ [2]

From [1]

$10^1 = 10 = k$

Substitute this into [2]

$2 = Alog_{10} (30)$

$A = \frac {2}{log_{10}(30)}$

$A = \frac {2} {log_{10}(2) + log_{10}(5) + log_{10}(3)}$

Correct me if I'm wrong but I believe you've made some incorrect assumptions

$\begin{align*} 1&=\text{A}\cdot\log_{10}{(k)}\\ 2&=\text{A}\cdot\log_{10}{(3k)}\\ \end{align*}$

Subtract second from the first...

$\begin{align*} 1&=\text{A}\cdot\log_{10}{(3k)}-\text{A}\cdot\log_{10}{(k)}\\ &=\text{A}\log_{10}{(3)}\\ \text{A}&=\frac{1}{\log_{10}{(3)}}\\ &=\log_{3}{(10)} \end{align*}$

Sub that into the first equation...

$\begin{align*} 1&=\log_{3}{(10)}\cdot \log_{10}{(k)}\\ &=\frac{\log_e{(10)}}{\log_e{(3)}}\cdot \frac{\log_e{(k)}}{\log_e{(10)}}\\ &=\log_{3}{(k)}\\ k&=3 \end{align*}$

Therefore $\text{A}=\log_3{(10)}$ and $k=3$

brightsky · « **Reply #6 on:** January 02, 2010, 08:01:33 pm »

Ahh yes yes yes!!! I forgot the A in [1]. Thanks for spotting that!!

tolga · « **Reply #7 on:** January 03, 2010, 01:55:42 pm »

may i ask what division you did there or what rule

tolga · « **Reply #8 on:** January 03, 2010, 02:06:00 pm »

for question 2 why did you divide by aren't you supposed to subtract

tolga · « **Reply #9 on:** January 03, 2010, 02:07:43 pm »

for the last questions step why did you do loge10/loge3(times) logek/loge10

GerrySly · « **Reply #10 on:** January 03, 2010, 05:57:21 pm »

Quote from: tolga on January 03, 2010, 02:06:00 pm

for question 2 why did you divide by aren't you supposed to subtract

I divided because it gets rid of $\text{A}$ easily, $\frac{\text{A}}{\text{A}}=1$

Quote from: tolga on January 03, 2010, 02:07:43 pm

for the last questions step why did you do loge10/loge3(times) logek/loge10

It's the change of base rule

$\log_{3}{(10)}=\frac{\log_e{(10)}}{\log_e{(3)}}$ , so you change both to base e so that the $\log_e{(10)}$ cancels out and you are just left with one log

tolga · « **Reply #11 on:** January 03, 2010, 06:28:40 pm »

for question 2 you divided becasue the bases Ae are the same, because ive never seen division is it possible to show with elimination or subtraction method

tolga · « **Reply #12 on:** January 03, 2010, 06:33:40 pm »

how did you get log3k what if you subbed 1/log10(3) then how would it turn out

GerrySly · « **Reply #13 on:** January 03, 2010, 09:44:45 pm »

Quote from: tolga on January 03, 2010, 06:33:40 pm

how did you get log3k what if you subbed 1/log10(3) then how would it turn out

It would turn out exactly the same actually... that is another application of the base change rule

$\begin{align*} \frac{1}{\log_{10}{(3)}}&=\frac{1}{\frac{\log_e{(3)}}{\log_e{(10)}}}\\ &=\frac{\log_e{(10)}}{\log_e{(3)}}\\ &=\log_{3}{(10)} \end{align*}$

Are you familiar with the change of base rule?

$\log_{a}{(b)}&=\frac{\log_e{(a)}}{\log_e{(b)}}$

Quote from: tolga on January 03, 2010, 06:28:40 pm

for question 2 you divided becasue the bases Ae are the same, because ive never seen division is it possible to show with elimination or subtraction method

brightsky used substitution I believe, the other methods I believe would be better if you did them yourself

"Give a man a fish and you feed him for a day. Teach him how to fish and you feed him for a lifetime." - Lao Tzu

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Author Topic: soving simulataneous equations for exponential functions? (Read 1121 times) Tweet Share

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