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!!(Punch)

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- Mar 19th 2008, 12:06 PMimstartingtohatethisconstant interest
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!!(Punch) - Mar 19th 2008, 12:15 PMMoo
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 - Mar 19th 2008, 12:19 PMimstartingtohatethis
i still dont get this...

- Mar 19th 2008, 12:29 PMMoo
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=$\displaystyle N+\frac{1.5}{100} N = N(\frac{100}{100}+\frac{1.5}{100}) = N(\frac{101.5}{100})$

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=$\displaystyle N1(\frac{101.5}{100}) = N(\frac{101.5}{100})^2$

So the general expression for the population at year n is :

$\displaystyle N (\frac{101.5}{100})^n$

And you want to know n such as the population at this year will reach 250,000 (or more).

So solve $\displaystyle 200,000 * (\frac{101.5}{100})^n > 250,000$

If we divide both sides by 200,000 , we have :

$\displaystyle (\frac{101.5}{100})^n > \frac{250000}{200000} = \frac{5}{4}$

Then use the logarithm :

$\displaystyle n \ln(\frac{101.5}{100}) > \ln(\frac{5}{4}) = 0.223$

$\displaystyle \ln(\frac{101.5}{100}) = 0.0149$

So $\displaystyle n > \frac{0.223}{0.0149}$

$\displaystyle n > 14.9$

As the number of years is an integer, the population will reach 250,000 within 15 years. - Mar 19th 2008, 12:45 PMSoroban
Hello, imstartingtohatethis!

Quote:

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: .$\displaystyle P \;=\;P_o(1 + r)^n$

. . where $\displaystyle P_o$ is the initial population,

. . $\displaystyle n$ is the number of years since 1990,

. . and $\displaystyle r$ is the rate of increase.

So we have: .$\displaystyle P \;=\;200,000(1.015)^n$

When does $\displaystyle P = 250,000$ ?

We have: .$\displaystyle 200,000(1.015)^n \;=\;250,000$

Divide by 20,000: .$\displaystyle 1.015^n \:=\:1.25$

Take logs: .$\displaystyle \ln(1.015^n) \;=\;\ln(1.25) \quad\Rightarrow\quad n\!\cdot\!\ln(1.015) \;=\;\ln(1.25)$

Therefore: .$\displaystyle n \;=\;\frac{\ln(1.25)}{\ln(1.015)} \;=\;14.98753167 \;\approx\;15$

The population will reach 250,000 in about 15 years . . . in 2005.