Question

The growth of bacteria makes it necessary to​ time-date some food products so that they will be sold and consumed before the bacteria count is too high. Suppose for a certain product the number of bacteria present is given by f(t)=600e0.1t​, where t is time in days and the value of​ f(t) is in millions. Find the number of bacteria present at each time. ​(a) 2 days ​(b) 3 days ​(c) 1 week

Answer 1

a) Substitute t = 2 into the formula

= 600e^0.1(2)

= 600e^0.2

= 732.84 million

Therefore, 732.84 million bacteria are present after 2 days

b) Substituting t = 4 days we get

= 600e^0.1(4)

= 600e0.4

= 895.09 million

Hence, 895.09 million bacteria are present after 4 days

c) 2 weeks = 14 days

Substituting t = 14 days we get

= 600e^0.1(14)

= 600e^0.4

= 2433.12 million

Thus, 2433.12 million bacteria are present after 2 weeks