Question

Harriet is cultivating a strain of bacteria in a petri dish. Currently, she has 103 bacteria in the dish. The bacteria divide every two hours such that the number of bacteria has doubled by the end of every second hour. How many bacteria will Harriet have in the dish at the end of 6 hours

Answer 1

Formula for growth or decay

R=
`R_(0)[1 \pm (r)/(100)]^t`

R= Final population of Bacteria

`R_(0)`= Initial Population

r = Rate of growth or decay

t= Time period

`R_(0)`= 103 bacteria

when, t=2

R becomes 103 × 2= 206 bacteria

Substituting these values in the formula of growth

206 = 103 ×
`(1+ (r)/(100))^2`

Dividing both sides by 103, we get

2 =
`(1+ (r)/(100))^2`

`(1+ (r)/(100))`= √2------(1)

`(1+ (r)/(100))`= 1.414

r = (1.414 -1) × 100= .414 × 100= 41.4%

Number of bacteria at the end of 6 hours = 103 ×
`(1+ (r)/(100))^6`= 103 ×
`[√(2)]^6` ------Using (1)

= 103 × 2³

=103 × 8

= 824→→bacteria that Harriet have in the dish at the end of 6 hours

Answer 2

Based on the conditions given above, the number of bacteria at any time t (in hours) is calculated by the equation,
at = (a1)(2^t/2)

where a1 is the initial number of bacteria and at is the number at any time t. Substituting the givens,
a6 = (103)(2^6/2) = 824

Thus, there are 824 bacteria after 6 hours.