Task 1
a. Let's assume the hourly growth rate of the bacteria is 3.
Hour | Number of Bacteria
-----|------------------
0 | 1
1 | 3
2 | 9
3 | 27
4 | 81
5 | 243
6 | 729
After 24 hours, the number of bacteria will be 3^24 = 282,429,536,481.
b. This table represents exponential growth because the number of bacteria is being multiplied by the growth rate (3) for each hour. The growth rate is constant, and each value in the table is obtained by multiplying the previous value by the growth rate.
c. Any nonzero number raised to the power of zero is equal to one because any number multiplied by one is equal to itself. In this example, raising the growth rate (3) to the power of zero is equivalent to multiplying by one, so the result is still 3^0 = 1.
d. The rule for this table can be represented as: N = 3^h, where N is the number of bacteria and h is the number of hours.
e. If we started with 100 bacteria, the rule would change to: N = 100 * 3^h, where N is the number of bacteria and h is the number of hours. The initial population is multiplied by the growth rate (3) raised to the power of the hours.
Task 2
a. Let's say the city is New York City, and its population on January 1st, 2020, was 8,398,748.
An exponential function to represent the city's population would be: y = 8,398,748 * e^(kx), where y is the population, x is the number of years that pass, and k is the growth rate constant.
b. Let's say the city is Los Angeles, and its population on January 1st, 2020, was 3,979,576.
An exponential function to represent the city's population would be: y = 3,979,576 * e^(kx), where y is the population, x is the number of years that pass, and k is the decay rate constant.
c. Both equations represent exponential growth/decay, as they involve the base of natural logarithms (e) raised to a power. The variables used in both equations are the population (y), the number of years that pass (x), and the growth/decay rate constant (k). The main difference is that one equation represents population growth (a) while the other represents population decline (b).
d. To find when the population of city (a) first exceeds that of city (b):
8,398,748 * e^(kx) > 3,979,576 * e^(kx)
Divide both sides by 3,979,576:
2.109 > e^(kx)
Take the natural logarithm of both sides:
ln(2.109) > ln(e^(kx))
ln(2.109) > kx
Solve for x:
x < ln(2.109) / k
Using data specific to the cities' growth rates, we can determine the value of k and compute the year when this inequality is true.
e. To find when the population of city (a) is at least twice the size of the population of city (b):
8,398,748 * e^(kx) >= 2 * 3,979,576 * e^(kx)
Divide both sides by 3,979,576:
2.109 >= e^(kx)
Proceeding similarly as in part (d), we can solve for x to determine the year when this inequality is true.