rate and sequences

[darkyadoo]

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 $r$ the rate (r:=5.2%) and $u_n$ the amount of the population after a delay of $n$ years.

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

Thanks for your help !

 

[Harry_the_cat]

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]

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!