The population of a Midwestern city decays exponentially. If the population decreased from 900,000 to 800,000 from 2003 to 2005, what will be the population in 2008?
I thought it was 7542 but it is wrong can someone please help me thank you
let the rate of decay be r, where r is a decimal
so letting 2003 correspond with t = 0 , then
2005 corresponds with t = 2
900,000 r^2 = 800,000
r^2 = 8/9
r = √(8/9) = appr .9428
So count = 9000(.0428)^t
in 2008 , t = 5
count = 900,000(.9428)^5 = appr 670,442
or 670,000 rounded to the nearest thousand
pop = 900000 * k^t
800000 = 900000 * k^2
8/9 = k^2 ... k = 2√2 / 3
p = 900000 * (2√2 / 3)^5
To find the population in 2008, we need to determine the growth rate of the population and apply it to the initial population in 2005.
The exponential decay formula is given by:
P(t) = P0 * e^(kt)
Where:
P(t) = population at time "t"
P0 = initial population
k = decay rate (negative value for decay)
t = time elapsed
First, we need to determine the decay rate "k". We can use the information given that the population decreased from 900,000 to 800,000 from 2003 to 2005.
Let's set up the equation for the given information:
P0 = 900,000
P(2005) = 800,000
Substituting the values into the formula:
800,000 = 900,000 * e^(k * 2)
Dividing both sides by 900,000:
800,000/900,000 = e^(2k)
Simplifying:
0.8889 = e^(2k)
To solve for "k", we can take the natural logarithm (ln) of both sides:
ln(0.8889) = ln(e^(2k))
Using the logarithmic rule ln(a^b) = b * ln(a):
ln(0.8889) = 2k * ln(e)
Since ln(e) = 1, we can simplify further:
ln(0.8889) = 2k
Calculating the value on the left side:
k ≈ ln(0.8889) / 2
Using a calculator, we find that k is approximately -0.0576.
Now, we can find the population in 2008, which is 3 years after 2005:
P(2008) = P0 * e^(kt)
Substituting the known values:
P(2008) = 800,000 * e^(-0.0576 * 3)
Calculating this expression, we find that the population in 2008 is approximately 739,702.
Therefore, the population in 2008 is 739,702, not 7,542 as you had initially calculated.
To find the population in 2008, we need to use the exponential decay formula. The formula for exponential decay is:
P(t) = P₀ * e^(kt)
Where:
P(t) = the population at time t
P₀ = the initial population
e = Euler's number (approximately 2.71828)
k = the decay constant
t = the time elapsed
In this case, we are given the initial population, P₀ = 900,000, and the population after 2 years, P(2) = 800,000.
Using these values, we can solve for the decay constant, k, by rearranging the equation:
P(t) = P₀ * e^(kt)
P(2) = 900,000 * e^(2k)
Dividing both sides of the equation by P₀:
P(2)/P₀ = e^(2k)
Substituting the given values:
800,000/900,000 = e^(2k)
Now, take the natural logarithm (ln) of both sides to isolate the exponent:
ln(800,000/900,000) = 2k
Using a calculator, evaluate the left side:
ln(800,000/900,000) ≈ -0.1053605
Divide both sides by 2:
k ≈ -0.05268025
Now that we have the decay constant, k, we can use it to find the population in 2008.
Using the formula:
P(t) = P₀ * e^(kt)
Substituting the values:
P(2008) = 900,000 * e^(-0.05268025 * (2008-2003))
Simplifying:
P(2008) = 900,000 * e^(-0.26340125)
Using a calculator:
P(2008) ≈ 783,031.24
Therefore, the population in 2008 is approximately 783,031.