need some help with a difference equation

- anonymous

need some help with a difference equation

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- anonymous

wats the question

- anonymous

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.

- anonymous

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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## More answers

- amistre64

recurrence equation tends to get better results... 'difference' equation is the same thing, but a less used term

- anonymous

alrightt

- amistre64

what was the original question? it seems like we are trying to jump into it halfway thru

- anonymous

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.

- anonymous

Maybe my whole difference equation is wrong....

- amistre64

it might be; 'btm' asks alot of these questions and may have some insight

- anonymous

what is btm?

- anonymous

I might be able to help

- amistre64

what is ncmtj?

- amistre64

imran is smart as well :)

- anonymous

alright thank you

- anonymous

thank you amistreo

- anonymous

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.

- anonymous

yn=8000-(.80n)800y_o ?

- anonymous

it should be 8000, no y_0

- anonymous

can you send me whole question

- anonymous

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.

- anonymous

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.

- anonymous

For my difference equation I have yn=〖.80〗_(yn-1) y0=8000

- anonymous

Okay, So I see you are solving for number of healthy trees. which is fine

- anonymous

yeah

- anonymous

Infected trees
y0=8000
y1=8000*(.8)
y2= 8000*(.8)^2

- anonymous

yn=8000*(.8)^2

- anonymous

*correction
yn=8000*(.8)^n

- anonymous

Total - Healthy trees= infected trees
8000-8000*(.8)^n=7560

- anonymous

-8000*(.8)^n=7560-8000
-8000*(.8)^n=-440
(.8)^n=-440/-8000
.8^n=.055
log[.8,.555]=n
n=12.99

- anonymous

ok. that does make much better sense. But how from 8000-8000(.80)^n, did u get 7560?

- anonymous

7560 was given in your problem
"the number of infected elm trees will be 7560"

- anonymous

thats right ok.

- anonymous

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.

- anonymous

your difference equation is
\[y_n=y_{n-1}*.8\]

- anonymous

alright that makes more sense now!!!

- anonymous

would y0 be 8000?

- anonymous

Yes

- anonymous

This is a difference equation for healthy trees

- anonymous

any other question?

- anonymous

that should be it. Thank you so much for your help.

- anonymous

np

- anonymous

hey imranmean, one more question. on the -8000(.80)n= -440, how did it come to Log[.8,.055]?

- anonymous

\[-8000*(.8)^n=7560-8000\]
\[-8000*(.8)^n=-440\]
\[(.8)^n={-440\over-8000}\]
.8^n=.055
log[.8,.555]=n
n=12.99

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