The population of Knoxville is 500,000 and is increasing at the rate of 3.75% each year. Approximately when will the population reach 1 million? Use an exponential model to solve the problem.
dp/dt = .0375 people/year
ln p = .0375 t + c
p = c e^(.0375 t)
when t = 0, p = 500,000
so
p = 500,000 e^(.0375 t)
so
2 = 1,000,000/500,000 = e^(.0375 t)
ln 2 = .0375 t
t = 18.5 years
To solve this problem using an exponential model, we can use the formula:
P(t) = P(0) * e^(rt),
where:
P(t) represents the population at time t,
P(0) represents the initial population,
e is Euler's number (approximately 2.71828),
r is the annual growth rate,
t is the time in years.
We are given:
P(0) = 500,000,
r = 3.75% = 0.0375.
We need to find t when P(t) = 1 million.
Substituting the given values into the formula, we have:
1,000,000 = 500,000 * e^(0.0375t).
To find t, we need to isolate it on one side of the equation. Let's divide both sides by 500,000:
2 = e^(0.0375t).
Next, we need to take the natural logarithm (ln) of both sides to cancel out the exponential function:
ln(2) = ln(e^(0.0375t)).
Using the logarithm rule, ln(e^x) = x, we can simplify the equation further:
ln(2) = 0.0375t.
Now, divide both sides by 0.0375 to solve for t:
t = ln(2) / 0.0375.
To solve this problem using an exponential model, we need to use the formula for exponential growth:
P(t) = P₀ * (1 + r)^(t)
Where:
P(t) is the population at time t
P₀ is the initial population
r is the growth rate (in decimal form)
t is the time in years
In this case, P₀ = 500,000, r = 0.0375 (which is 3.75% expressed as a decimal), and we want to find the value of t when P(t) is 1,000,000.
So, the equation becomes:
1,000,000 = 500,000 * (1 + 0.0375)^t
Now, let's solve for t. Divide both sides of the equation by 500,000 to isolate the exponential term:
2 = (1.0375)^t
To solve for t, we need to take the logarithm of both sides. Let's use the natural logarithm (ln) because it's a common choice in exponential growth problems:
ln(2) = ln((1.0375)^t)
Using the logarithm power rule, we can bring down the exponent:
ln(2) = t * ln(1.0375)
Finally, divide both sides of the equation by ln(1.0375) to isolate t:
t = ln(2) / ln(1.0375)
Using a calculator, we find that t ≈ 30.02 years.
Therefore, the population of Knoxville will reach 1 million approximately 30.02 years from now.