Question

use the formula P=Ie^kt. a bacterial culture has an initial population of 10000 . If its population declines to 3000 in 8 hours , what will it be at the end of 10 hours?

Answer 1

P(0)=10000, substitute this info into the equation P=Ie^kt,10000=Ie^0 implies 10000=I,
updating the equation, P(t)=10000e^ktP(8)=3000, substitute this into the equation above,3000=10000e^8k,0.3=e^8k,put natural logarithm at both right hand side and left hand side of this equation,ln 0.3 = 8k,1/8   ln 0.3 = k-0.1505=k
now we have P(t)=10000e^(-0.1505t).
just plug in t=10 into the equation 

Answer 2

Answer:

27

Step-by-step explanation:

Using the formula P =Ie^kt.

If bacterial culture has an initial population of 10,000, the equation becomes,

10,000 = Ie^k(0)... 1 where t = 0

10000 = I

If its population declines to 3000 in 8 hours, P = 3000 and t = 8hours

This gives;

3000 = Ie^8k... 2

From equation 2, we take natural logarithm of both sides to have;

ln3000 = I lne^8k

ln3000 = 8Ik

k = ln3000/8I

k = ln3000/8(10000) since I is 10000

k = ln3000/80000

k = 8/80000

k = 0.0001

At the end of 10hours, the population will become;

Since P = Ie^kt

I = 10000 k = 0.0001 and t = 10

P = 10000e^(0.0001×10)

P = 10000e^0.001

P = 10000×0.0027

P = 27

The population of the bacteria will have reduced to about 27population after 10hours