The rodent population in a city is currently estimated at 30,000 and is growing according to the Malthusian model. If it is expected to double every 9 years, when will the population reach one million? (Round your answer to one decimal place.)
No idea what the Malthusian model is, but I will do it my way
P = 30,000 e^(kt) , where k is constant and t is years
given:when t = 9, P = 60,000
30000 e^(9k) = 60000
e^(9k) = 2
9k = ln2
k = ln2/9
P = 30,000 e^((ln2/9)t)
when is P = 1,000,000 ?
30,000 e^((ln2/9)t) = 1,000,000
e^((ln2/9)t) = 100/3
ln2/9 t = ln100 - ln3
t = 9(ln100 - ln3)/ln2 = 45.5 years
alternate way:
since there is doubling effect, we can use base 2 in our exponential term
P = 30,000 (2)^(t/9)
30,000 2^(t/9) = 1,000,000
2^(t/9) = 100/3
log (2^(t/9)) = log(100/3)
t/9 log2 = log100 - log3
t/9 = (2 - log3)/log2
t = 9(2 - log3)/log2 = 45.5 years
Thank you so much for your help!
To find out when the population will reach one million, we can use the Malthusian model formula:
P(t) = P(0) * e^(r*t)
where:
P(t) represents the population at time t
P(0) represents the initial population
r represents the growth rate
t represents the time in years
Given:
P(0) = 30,000 (initial population)
r = ln(2)/9 (growth rate, derived from "doubling every 9 years")
P(t) = 1,000,000 (target population)
Now, substitute the values into the formula:
1,000,000 = 30,000 * e^(ln(2)/9 * t)
To solve for t, we need to isolate the variable t. Divide both sides of the equation by 30,000:
(1,000,000 / 30,000) = e^(ln(2)/9 * t)
33.33 = e^(ln(2)/9 * t)
Next, take the natural logarithm of both sides of the equation to remove the exponential:
ln(33.33) = ln(e^(ln(2)/9 * t))
ln(33.33) = ln(2)/9 * t
Use the properties of logarithms to simplify the equation:
t = (9 * ln(33.33)) / ln(2)
Using a calculator, we can solve for t:
t ≈ 63.0 years (rounded to one decimal place)
Therefore, the population is expected to reach one million after approximately 63.0 years.
To find out when the rodent population will reach one million, we can use the Malthusian model formula:
P(t) = P₀ * e^(rt)
Where:
P(t) is the population at time t
P₀ is the initial population
e is the base of the natural logarithm (approximately 2.71828)
r is the growth rate
t is the time in years
In this case, the initial population (P₀) is 30,000 and it is expected to double every 9 years, so the growth rate (r) can be calculated using the formula:
r = ln(2) / t
Let's substitute these values into the formula and solve for t:
P(t) = P₀ * e^(rt)
1,000,000 = 30,000 * e^((ln(2)/9)t)
Now, divide both sides of the equation by 30,000:
1,000,000 / 30,000 = e^((ln(2)/9)t)
Simplifying further:
33.333 = e^((ln(2)/9)t)
To solve for t, take the natural logarithm of both sides:
ln(33.333) = (ln(2)/9)t
Now, solve for t:
t = (ln(33.333) * 9) / ln(2)
Calculating the values:
t ≈ (3.505 * 9) / 0.6931
t ≈ 31.545 / 0.6931
t ≈ 45.522
Therefore, the rodent population is estimated to reach one million in approximately 45.5 years.