Question

A bacterial colony starts with 120 cells and quadruples in size each day. Write an equation that relates the population of cells in this colony (P) at the start of each day and the number of days (d). Find the population on day 8.

Answer 1

P = 120*(
`2^(2d-2)`)

P(8) = 1, 966, 080

Explanation:

Since the population quadruples each day, the population for the subsequent day would be 4*(population of the previous day).

Thus, the evolution of the population value takes the form of a geometric progression, with a common ratio, r = 4

The n-th term of a geometric progression is given by:

`a_(n)` =
`ar^(n-1)` (1)

Where a is the 1st term of the progression.

From (1), our population would generally take the form:

`P_(d)` =
`P_0r^(d-1)` (2)

In this case, the initial value (1st term)
`P_(0)` = 120.

So putting r and
`P_(0)` into (2):

P(d) = 120*(
`4^(d-1)`)

Noting that 4 = 2²:

P(d) = 120*(
`2^(2(d-1))`)

P(d) = 120*(
`2^(2d-2)`)

FOR d = 8:

P(d) = 120*(
`2^(2d-2)`)

becomes:

P(8) = 120*(
`2^(2(8)-2)`)

P(8) = 120*(
`2^(16-2)`)

P(8) = 120*(
`2^(14)`)

P(8) = 120*(16,384)

P(8) = 1, 966, 080