Question

at the start of the experiment there are 400 bacteria if the bacteria follow an exponential growth pattern with r=0.03 what will be the population after 6 hours? how long will it take for the population to double?​

Answer 1

1. answer. 400+6 is the answer

Answer 2

Given:

Initial no. of bacteria = 400

Rate of growth = 0.03

Time = 6 hours

To find:

Population of bacteria after 6 hours.

Time taken to double the population.

Solution:

Let $A=A_{0}e^{rt}$

where, $A_{0}$ is the initial population, $r$ is the rate of growth and $t$ is the time taken.

Substituting the values,

$A=400e^{0.03*6}$

$A=400*1.2$

$A=480$

∴ Population of bacteria after 6 hours = 480

When population doubles, $A = 800, A_{0}=400, r=0.03$

$A=A_{0}e^{rt}$

$800=400e^{0.03t}$

$\frac{800}{400}=e^{0.03t}$

$200=e^{0.03t}$

$ln(200)=0.03t$

$5.3=0.03t$

$t=176.67$ ≈ $177$ hours

∴ Time taken for the population to double is 177 hours.

Population of bacteria after 6 hours = 480

Time taken for the population to double is 177 hours.