the population of a city is given by 3000e^(kt), where t = 0 corresponds to the year 1960. in 1950 the population was 2400. Find k to the thousandth place, and predict the population in 2015.
using the point (-10,2400), we have
3000 e^(-10k) = 2400
so, k = 1/10 ln 1.25 = 0.022
Now you can use the function as needed
so do it
e^0 = 1 so population was 3000 in 1960
when t = - 10, p = 2400
3000 e^-10k= 2400
e^-10 k = 24/30 = 12/15 = 0.8
ln e^-10 k = -10 k = ln 0.8 = -.223
k = .223/10 = 0.0223
so
p = 3000 e^0.0223 t
2015 - 1960 = 55 = new
so put 55 in for t
if e^-10k = 4/5, then e^10k = 5/4 = 1.25
You are welcome.
@oobleck So I assume 1/10 is 1 year, but where does ln 1.25 come in from? If I follow correctly, you should end up with 600 = e^(-10k). Correct me if I made a mistake.
I understand now. I was doing 2400 - 3000. Had I don't division and followed through, I would have seen it. Thank you both!
To find k, we can use the given information that the population in 1950 was 2400. This means that t = -10 (because t = 0 corresponds to the year 1960, and 1950 is 10 years before 1960).
Substituting the known values into the given equation:
2400 = 3000e^(k*(-10))
Simplifying the equation:
0.8 = e^(-10k)
To solve for k, we need to take the natural logarithm (ln) of both sides of the equation:
ln(0.8) = -10k
Now, we can solve for k:
k = ln(0.8) / -10
Using a calculator, we find k ≈ -0.022.
To predict the population in 2015, we need to find the value of t when the year is 2015. Since t = 0 corresponds to the year 1960, the population in 2015 is when t = 55 (because 2015 is 55 years after 1960).
Substituting the known values into the equation:
Population in 2015 = 3000e^(k*55)
Population in 2015 = 3000e^(-0.022*55)
Using a calculator, we find that the predicted population in 2015 is approximately 1873.