Algebera HELPP!!. Urgent

Question

I have done the work but the teacher told me it's wrong can someone please help me with this? With steps? I am confused

When a cell cycle for an unknown organism starts, its cells start replicating and are known to triple every four hours. let's suppose initially there were 50 cells.

(a))How many cells are there after 12 hours?

(b) Please use an exponential function (C(t)=C(subscript of 0)e^rt/ C0e^rt) to represent the cells after t hours, and solve for C0 and r.

(c) Estimate how many cells will there be after 33 hours. answer is asked to be in the most exact representation (i.e., not a decimal).

Answer 1

Hi Josh,

Parts a and b can be done together. Write down what you know:

a.

c₀=50–the number of cells you have at the start.

c₄=150–since you know the number of cells triples every four hours

Now use the equation your instructor specified in part b:

C(t)=C₀e^rt

Now let’s break down what we know and solve for r:

t=4

C(t)=C(4)=150

e=Exponential Function, valued at about 2.718

150=50e^4r

Divide both sides by 50 to isolate r:

3=e^4r

Take the natural log of both sides to eliminate the exponent:

ln (3)=4r [ln(e)]

Recall ln(e)=1

ln(3)=4r

r=ln(3)/4

In decimal form, ln(3)/4 is about 0.275. Now that we know r, we can substitute values into our initial equation when t=12 to get the total cells after 12 hours.

C(t)=C₀e^rt

C₀=50

e=e=about 2.718

t=12

r=ln (3)/4, about 0.275

Substituting:

C(12)=50e^12[ln(3)/4]

C(12)=1350 cells

b. R=ln(3)/4, from part a

C₀=50 cells

n₀ values are always the number of cells, dollars, etc. you start off with, it is also sometimes written as n_i where I stands for initial.

c. I can’t quite get this one, but to start:

C(t)=C₀e^rt

C₀=50

e=e, about 2.718

r=ln(3)/4

t=33

C(33)=50e^33[ln(3)/4]

C(33)=431738 cells

That’s still an approximation, though, and I’m not sure how to simplify the last equation. If someone else who knows more about logarithms than me wants to step in, they are welcome to. I hope this helps you.

Answer 2

Lets start by stating that the cell population triples every 4 hours and that we start with only 1 cell:

3^(t/4) does this precisely since one cell becomes 3 after 4 hours.

A. If the initial number of cells is a, then we have

f(t) = a × 3^(t/4)

If the initial cell population is 50. Then after 12 hours,

f(t) = 50 × 3^(12/4) = 50 × 3^(3) = 50 × 27 = 1,350 cells

B. To convert to base e, note that

b^t = e^(ln(b^t)) = e^(t×ln(b))

In our case,

C(t) = C₀ × e^{((t/4)×ln(b))} = C₀ × e^{((ln(3)/4)×t)}

r = ln(3)/4

C. After 33 hours,

C(t) = 50× e^{((ln(3)/4)×33)}

Or

50 × 3^(33/4)

You have to convert to a decimal after this point.

The calculated value of the above expressions at t=33 is 431,738 cells (discarding the associated fraction).