The population of a certain town grows from 10,000 in 1982 to 35,000 in 1995. At this rate, in what year will the population reach 300,000?
Tip* Population growth formula:
P = P(subzero)multiplied by e^(k)(t)
p = po e^(kt)
10,000 = 10,000 e^0 if t = 0 at 1982
1995 - 1982 = 13
35,000 = 10,000 e^(k*13)
ln(3.5) = 13 k
k = 1.253/13 = .09637
300,000 = 10,000 e^(.09637 t)
ln 30 = .09637t
t = 35.3
1982 + 35.3 = 2017 almost there :)
To solve this problem, we can use the population growth formula:
P = P₀ * e^(kt)
where:
P = population at a given time (t)
P₀ = initial population
k = growth rate constant
t = time (in years)
We can start by finding the value of the growth rate constant (k). We are given the initial population in 1982 (P₀ = 10,000) and the population in 1995 (P = 35,000). Let's calculate k:
35,000 = 10,000 * e^(k*13)
Dividing both sides of the equation by 10,000:
3.5 = e^(13k)
To isolate e^(13k), we take the natural log (ln) of both sides:
ln(3.5) = 13k
Now we can solve for k:
k = ln(3.5) / 13
Using a calculator, we find that k ≈ 0.097.
Now we can substitute the values into the population growth formula to find the year when the population will reach 300,000:
300,000 = 10,000 * e^(0.097t)
Dividing both sides of the equation by 10,000:
30 = e^(0.097t)
To isolate e^(0.097t), we take the natural log of both sides:
ln(30) = 0.097t
Finally, we solve for t by dividing both sides of the equation by 0.097:
t ≈ ln(30) / 0.097
Using a calculator, we find that t ≈ 26.8.
Therefore, the population will reach 300,000 approximately in the year 1982 + 26.8 = 2009.8. Rounding to the nearest year, the year will be 2010.
To find the year when the population will reach 300,000, we can use the population growth formula P = P₀ * e^(kt), where P is the final population, P₀ is the initial population, e is a constant approximately equal to 2.71828, k is the growth rate, and t is the time in years.
First, let's find the growth rate, k. We can use the following equation to solve for k:
k = (ln(P/P₀)) / t
ln denotes the natural logarithm function.
Substituting the given values, P = 35,000, P₀ = 10,000, and t = 1995 - 1982 = 13 years, we can calculate k:
k = (ln(35,000/10,000)) / 13
Now, we have the growth rate. We can rewrite the population growth formula as:
P = P₀ * e^(kt)
Substituting the known values, P₀ = 10,000, P = 300,000, and k = (ln(35,000/10,000)) / 13, we can solve for t:
300,000 = 10,000 * e^(13 * k)
Dividing both sides by 10,000:
30 = e^(13k)
Take the natural logarithm of both sides:
ln(30) = ln(e^(13k))
Using the properties of logarithms, we can bring down the exponent:
ln(30) = 13k * ln(e)
Since ln(e) is equal to 1, the equation simplifies to:
ln(30) = 13k
Now, divide both sides by 13:
k = ln(30) / 13
With the value of k, we can now solve for t when the population reaches 300,000. Let t be the number of years from 1982:
300,000 = 10,000 * e^((ln(30) / 13) * t)
Divide both sides by 10,000:
30 = e^((ln(30) / 13) * t)
Take the natural logarithm of both sides:
ln(30) = (ln(30) / 13) * t * ln(e)
Since ln(e) is equal to 1, the equation simplifies to:
ln(30) = (ln(30) / 13) * t
Now, divide both sides by (ln(30) / 13):
t = ln(30) / (ln(30) / 13)
Calculate the expression on the right side:
t = 13
Therefore, the population will reach 300,000 in the year 1995 + 13 = 2008.