need some help with a difference equation

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need some help with a difference equation

Mathematics
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wats the question
yn=8000-(.80n)800o. Trying to solve for N. I know the answer I just need help with finding out how to get N alone and such. The answer should be like 13.
Im not sure if I need to divide by another number. I have another number and its 7560. Do I need to put the 7560 where the yn is

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Other answers:

recurrence equation tends to get better results... 'difference' equation is the same thing, but a less used term
alrightt
what was the original question? it seems like we are trying to jump into it halfway thru
Ok, it says to use the solved difference equation to find in what calender year, the number of infected elm trees will be 7560. You must solve the solved difference equation: you may not guess and check. Get an equation for n and solve it.
Maybe my whole difference equation is wrong....
it might be; 'btm' asks alot of these questions and may have some insight
what is btm?
I might be able to help
what is ncmtj?
imran is smart as well :)
alright thank you
thank you amistreo
I can send you more information about the problem to maybe be able to help you more. There is like five questions to be answered.
yn=8000-(.80n)800y_o ?
it should be 8000, no y_0
can you send me whole question
yes.Suppose a region has 8000 healthy elm trees. In 1950, bark beetles harboring the fungus arrive in the region. In each subsequent yera, the number of elm trees that contract the disease is 20% of the number of healthy trees at the end of the previous year. Once a tree is infected, it stays infected. Find the difference equation.
Also, to solve the difference equation, and use the solved difference equatoin to find in what calender year, the number of infected elm trees will be 7560. You must solve the solved difference equation.
For my difference equation I have yn=〖.80〗_(yn-1) y0=8000
Okay, So I see you are solving for number of healthy trees. which is fine
yeah
Infected trees y0=8000 y1=8000*(.8) y2= 8000*(.8)^2
yn=8000*(.8)^2
*correction yn=8000*(.8)^n
Total - Healthy trees= infected trees 8000-8000*(.8)^n=7560
-8000*(.8)^n=7560-8000 -8000*(.8)^n=-440 (.8)^n=-440/-8000 .8^n=.055 log[.8,.555]=n n=12.99
ok. that does make much better sense. But how from 8000-8000(.80)^n, did u get 7560?
7560 was given in your problem "the number of infected elm trees will be 7560"
thats right ok.
In my book the normal difference equations are written as for example, yn=-.8yn-1 +9, yo=10. Do I need to put any of those into this new difference equation. The equation seems bare.
your difference equation is \[y_n=y_{n-1}*.8\]
alright that makes more sense now!!!
would y0 be 8000?
Yes
This is a difference equation for healthy trees
any other question?
that should be it. Thank you so much for your help.
np
hey imranmean, one more question. on the -8000(.80)n= -440, how did it come to Log[.8,.055]?
\[-8000*(.8)^n=7560-8000\] \[-8000*(.8)^n=-440\] \[(.8)^n={-440\over-8000}\] .8^n=.055 log[.8,.555]=n n=12.99

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