rate and sequences

darkyadoo

New member
Joined
Aug 18, 2021
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34
Good afternoon,

The town of Wattle has a population that is decreasing steadily.

It is found that the population is decreasing at a rate of 5.2% per year.

How long will it take for the population to be beneath 46% of what it is now?

I would like you to confirm that my result is true on contrary to the one given by the correction.

we call [imath]r[/imath] the rate (r:=5.2%) and [imath]u_n[/imath] the amount of the population after a delay of [imath]n[/imath] years.

So after n years we have [imath]u_n=u_{n-1}(1-r)[/imath], so we deduce that [math]u_n=u_0(1-r)^n[/math]
Since we work out [imath]n[/imath] so that the population is beneath 46%, once that is equated that gives [imath]u_0(1-0.46)=u_0(1-r)^n[/imath] therefore we have :[math]u_0(1-0.46)=u_0(1-r)^n\iff \ln 0.54=n\ln 0.948\iff n=\dfrac{\ln 0.54}{\ln 0.948}=11.53[/math], we finally deduce the answer : [math]\boxed{n=12}[/math]
The correction gives n=15

Thanks for your help !
 

Harry_the_cat

Elite Member
Joined
Mar 16, 2016
Messages
3,133
You are incorrect. n=15 is correct.
Your calculation is based on a 46% reduction in the population. That is the population will be below 54% of what it is now.
There is no need to subtract 0.46 from 1.
You will then have: ln 0.46 = n ln 0.948 to solve which yields n=15.
 

darkyadoo

New member
Joined
Aug 18, 2021
Messages
34
You are incorrect. n=15 is correct.
Your calculation is based on a 46% reduction in the population. That is the population will be below 54% of what it is now.
There is no need to subtract 0.46 from 1.
You will then have: ln 0.46 = n ln 0.948 to solve which yields n=15.
Thanks !! I need to read 7 times the question!:)
 
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