In 1990 the population of a city was about 200,000. If the population increases at a constant rate of 1.5% per year, in what year is the population projected to reach 250,000?
Please help im going crazy!!
Hello,
Consider the function x(t) the number of inhabitants and t the number of years.
For the first year, 200000 becomes 200000*(100+1.5)/100
For the second year, the population becomes 200000*(100+1.5)/100 *(100+1.5)/100 = 200000*((100+1.5)/100)²
And so on... Each year, you multiply the number of inhabitants by (100+1.5)/100 (corresponds to an augmentation of 1.5%)
So x(t)=200000*(101.5/100)^t
And you want t such as x(t)>250,000
Hence the problem is to solve 200000*(101.5/100)^t > 250000
Well,
The initial population is at N=200,000 inhabitants
The year following, it increases of 1.5%, which means that to N, we add 1.5% of N
The population will then be N1=
The second year, it increases again of 1.5%, which means that to N1, we add 1.5% of N1
With the same scheme, we got N2=
So the general expression for the population at year n is :
And you want to know n such as the population at this year will reach 250,000 (or more).
So solve
If we divide both sides by 200,000 , we have :
Then use the logarithm :
So
As the number of years is an integer, the population will reach 250,000 within 15 years.
Hello, imstartingtohatethis!
You're expected to know the Exponential Growth formula.In 1990 the population of a city was about 200,000.
If the population increases at a constant rate of 1.5% per year,
in what year is the population projected to reach 250,000?
For this problem it is: .
. . where is the initial population,
. . is the number of years since 1990,
. . and is the rate of increase.
So we have: .
When does ?
We have: .
Divide by 20,000: .
Take logs: .
Therefore: .
The population will reach 250,000 in about 15 years . . . in 2005.