Pretty difficult problem, but that’s why I’m here.
First we need to identify what we’re looking for, which is t. So now plug 450k into equation and solve for t.
450000 = 250000e^0.013t
Now to solve this, we need to remember this rule: if you take natural log of e you can remove x from exponent. And natural log of e is 1.
Basically ln(e^x) = xln(e) = 1*x
So knowing this first we need to isolate e
450000/250000 = e^0.013t
1.8 = e^0.013t
Now take natural log of both
Ln(1.8) = ln(e^0.013t)
Ln(1.8) = 0.013t*ln(e)
Ln(1.8) = 0.013t * 1
Now solve for t
Ln(1.8)/0.013 = t
T= 45.21435 years
Now just to check our work plug that into original equation which we get:
449999.94 which is basically 500k (just with an error caused by lack of decimals)
So our final solution will be in the 45th year after about 2 and a half months it will reach 450k people.

6 0

2 years ago

The population of a town increases by 4 % every year . In the last year the population of the town was 110000.

TEA [102]

Answer:

114,400 and 118,976.

Step-by-step explanation:

Let's represent this using a function. Let's let:

$P(y)$ represent the total population, and $y$ represent the total number of years since last year.

We know that the population grows by 4% every year or 0.04 every year. This is exponential growth. We can write this as:

$P(y)=110000(1.04)^y$

Note that the coefficient is 110000 because that is the year we are starting with. Also, note that it is 1.04 because we are essentially adding .04 to the original population.

Anyways, to find the present population, set y equal to 1 (because 1 year after last year is the present year)

$P(1)=110000(1.04)^1=114400$