In an experiment, a certain colony of bacteria initially has a population of 50,000. A reading is taken every 2 hours, and at the end of every 2-hour interval, there are 3 times as many bacteria as before. a. Write a recursive definition for A(n), the number of bacteria present at the beginning of the nth time period. b. At the beginning of which interval are there 1,350,000 bacteria present?

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skyluke89 skyluke89 · Answer 1 · 2020-02-24T12:21:26+01:00

Answer:

1) Recursive definition: [tex]p_n = (50,000)3^n[/tex]

2) At the beginning of the 4th interval

Explanation:

1)

The initial population of the bacteria at time zero is

[tex]p_0 = 50,000[/tex]

Here we are told that the reading is taken every two hours; we call this time interval "n", so

[tex]n=2 h[/tex]

And also, after every time interval n, the number of bacteria has tripled.

This means that when n = 1,

[tex]p_1 = 3 p_0[/tex]

And when n=2,

[tex]p_2 = 3 p_1 = 3(3p_0)=9 p_0[/tex]

Applied recursively, we get

[tex]p_n = 3^n p_0[/tex]

And substituting p0,

[tex]p_n = (50,000)3^n[/tex] (1)

2)

Here we want to find at the beginning of which interval there are

[tex]p=1,350,000[/tex]

bacteria.

This means that we can rewrite eq.(1) as

[tex]1,350,000=(50,000)3^n[/tex]

By simplifying,

[tex]27=3^n[/tex]

Which means that

[tex]n=3[/tex]

However, this means that the number of bacteria is 1,350,000 after 3 time intervals; therefore, at the beginning of the 4th interval.