Question

A type of bacteria has a very high exponential growth rate of 80% every hour. If there are 10 bacteria, determine how many will be in 5 hours,1 day, and 1 week

Answer 1

There are 189 bacteria in 5 hours

There are 13382588 bacteria in 1 day

There are $10(1.8)^{168}$ bacteria in 1 week

Solution:

Given that,

A type of bacteria has a very high exponential growth rate of 80% every hour

There are 10 bacteria

The increasing function is given as:

$y = a(1+r)^t$

Where,

y is future value

a is initial value

r is growth rate

t is time period

From given,

a = 10

$r = 80 \5 = \frac{80}{100} = 0.8$

Determine how many will be in 5 hours

Substitute t = 5

$y = 10(1 + 0.8)^5\\\\y = 10(1.8)^5\\\\y = 10 \times 18.89568\\\\y \approx 188.96$

y = 189

Thus, there are 189 bacteria in 5 hours

Determine how many will be in 1 day ?

1 day = 24 hours

Substitute t = 24

$y = 10(1 + 0.8)^{24}\\\\y = 10(1.8)^{24}\\\\y = 10 \times 1338258.84\\\\y = 13382588.45\\\\y \approx 13382588$

Thus, there are 13382588 bacteria in 1 day

Determine how many will be in 1 week

1 week = 168

Substitute t = 168

$y = 10(1 + 0.8)^{168}\\\\y = 10(1.8)^{168}$

Thus there are $10(1.8)^{168}$ bacteria in 1 week