"A Bacteria Culture Grows with Constant Relative Growth Rate. A micro organism culture Grows with consistent Relative boom rate. The micro organism count number become 400 after 2 hours and 25,six hundred after 6 hours. a) What is the relative growth rate? Express your answer as a percentage. Using this formula P(t)=P_0e^kt t=time k=growth rate p_0=initial amount How do I calculate a formula with the information I have? b) What was the initial size of the culture? c) Find an expression for the number of bacteria after t hours d) Find the number of cells after 4.5 hours e) Find the rate of growth after 4.5 hours f) When the population reach 50,000? that is a self look at query. i've Calculus II subsequent semester and sincerely would like to have a clear method as to a way to solve a hassl

blogswput

blogswput

Answered question

2022-09-12

A Bacteria Culture Grows with Constant Relative Growth Rate.
A micro organism culture Grows with consistent Relative boom rate. The micro organism count number become 400 after 2 hours and 25,six hundred after 6 hours.
a) What is the relative growth rate? Express your answer as a percentage.
Using this formula P ( t ) = P 0 e k t
t = t i m e
k = g r o w t h   r a t e
p 0 = i n i t i a l   a m o u n t
How do I calculate a formula with the information I have?
b) What was the initial size of the culture?
c) Find an expression for the number of bacteria after t hours
d) Find the number of cells after 4.5 hours
e) Find the rate of growth after 4.5 hours
f) When the population reach 50,000?
that is a self look at query. i've Calculus II subsequent semester and sincerely would like to have a clear method as to a way to solve a hassle inclusive of this.

Answer & Explanation

kappastud98u

kappastud98u

Beginner2022-09-13Added 10 answers

So let t be the time in hours.
You know two things:
P ( 2 ) = P 0 e 2 k = 400
P ( 6 ) = P 0 e 6 k = 25600
This means you have two equations and two unknowns ( P 0 and k). To solve for k, you can divide the second equation by the first:
P 0 e 6 k P 0 e 2 k = 25600 400
e 4 k = 64
( e k ) 4 = 64 = 2 6
e k = 2 2 = 2 1.5
ln ( e k ) = ln 2 1.5
k = 3 ln 2 2 1.0397
Now that we know the value of k, we can substitute it into the first of our original two equations to find the value of P 0 :
P 0 e 2 k = 400
P 0 e 3 ln 2 = 400
P 0 ( e ln 2 ) 3 = 400
8 P 0 = 400
P 0 = 50
Note that P0 is the same thing as P(0), the bacteria count at time t=0 hours.

Do you have a similar question?

Recalculate according to your conditions!

Ask your question.
Get an expert answer.

Let our experts help you. Answer in as fast as 15 minutes.

Didn't find what you were looking for?