The online program at a certain university had an enrollment of 580 students at its inception and an enrollment of 1725 students 3 years later

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The online program at a certain university had an enrollment of 580 students at its inception and an enrollment of 1725 students 3 years later. Assume that the enrollment increases by the same percentage per year. a) Find the exponential function E that gives the enrollment t years after the online program's inception. b) Find E(11), and interpret the result. c) When will the program's enrollment reach 5500 students?

Let E0 be the number of students that enrolled at inception

E(t) be the number of students that enrolled at any time t

At inception i.e at t = 0, E0 = 580

The exponential function E that gives the enrollment at inception will be expressed as;

E(t) = E0e^kt (note that the exponential power is positive since number of students keep increasing yearly).

At inception:

E(0) = 580e^k(0)

E(0) = 580(1)

E(0) = 580

If after 3years, enrollment increases to 1725 students i.e at t = 3, E(t) = 1725. The equation becomes;

E(3) = P0e^kt

E(3) = 580e^3k

1725 = 580e^3k

Make k the subject of the formula:

e^3k = 1725/580

e^3k = 2.9741

Apply ln to both sides of the equation

ln(e^3k) = ln(2.9741)

3k = 1.08995

k = 1.08995/3

k = 0.3633

The exponential function E that gives the enrollment t years after the online program's inception will be expressed as;

E(t) = E0e^kt

Given E0 = 580 and k = 0.3633

E(t) = 580e^0.3633t

b) To get E(11), we will substitute t = 11 into the resulting expression in (a) above to have;

E(t) = 580e^0.3633t

E(11) = 580e^0.3633(11)

E(11) = 580e^3.9965

E(11) = 580×54.4074

E(11) = 31,556.287

E(11) ≈ 31,556 students

This means that about 31,556students enrolled 11years later.

c) To know when the program's enrollment reach 5500 students, we will set E(t) to 5500, E0 to 580 and k to 0.3633 and calculate the value of t

From the equation E(t) = 580e^0.3633t

5500 = 580e^0.3633t

e^0.3633t = 5500/580

e^0.3633t = 9.4828

Apply ln to both sides

ln(e^0.3633t) = ln(9.4828)

0.3633t = 2.2495

t = 2.2495/0.3633

t = 6.19

t ≈ 6years

This means that the program's enrollment will reach 5500 students 6 years later.

it’s supposed to be -4+45=41

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