Question

A cell of some bacteria divides into two cells every 10 minutes.The initial population is 3 bacteria. (a) Find the size of the population after t hours (function of t) (b) Find the size of the population after 7 hours. # Preview | Preview (c) When will the population reach 21? t42 Preview

Answer 1

Answer:

(a) $P_{t}=3(2)^{6t}$

(b) $3(2)^{42}$

(c) 28.07 minutes

Step-by-step explanation:

A cell of some bacteria divides itself into 2 cells in every 10 minutes and initial population of the bacteria was 3.

That means sequence formed will be 3, 6, 12, 24............

We can easily say that this sequence is a geometric sequence having common ratio (r) = $\frac{T_{2}}{T_{1}}=\frac{6}{3}$

r = 2

Now we know the explicit formula of a geometric sequence is given by

$P_{t}=P_{0}(r)^{\frac{60t}{10}}=P_{0}(r)^{6t}$

Where a = Initial population = 3 bacteria

r = common ratio = 2

and t = time in hours

So explicit formula will be $P_{t}=3(2)^{6t}$

(a) Now we have to calculate the size of population after t hours

$P_{t}=3(2)^{6t}$

(b) We have to find the size of population after 7 hours or 420 minutes

$P_{t}=3(2)^{6\times7}$

= $3(2)^{42}$

After 7 hours bacteria population will be $3(2)^{42}$

(c) Time to reach population as 21

By the explicit formula

$21=3(2)^{6t}$

$2^{6t}=\frac{21}{3}=7$

Now we take log on both the sides of the equation

$log(2^{6t})=log(7)$

6t log2 = log 7

6t(0.301) = 0.845

t(1.806) = 0.845

t = $\frac{0.845}{1.806}=0.468$ hours

Or t = 0.468×60 = 28.07 minutes

Therefore, after 28.07 minutes bacterial population will be 21