# Thread: Exponential function word problems?

1. ## Exponential function word problems?

I really don't enjoy word problems. Can you give me a hand with the following? Once it's in an equation I'm pretty good..

5. For a biology experiment, the number of cells present is 1000. After 4 hours, the count is estimated to be 256000. What is the "doubling period" of the cells?
* How do you derive an equation from this? I tried and got x= 1/2 as my answer. Am i anywhere close?

6. A certain bacteria strain divides every 0.5 hours. If 500 bacteria are present in a culture, how many will be there
a) After 3 hours?
B) After 6 hours?
C) After n hours?

10. A certain strain of yeast cell doubles every 20 minutes. If there were 350 initially, how many will there be in 3 hours?

Again guys, I could really use the help.
Thanks so much!

2. Originally Posted by brittt--
I really don't enjoy word problems. Can you give me a hand with the following? Once it's in an equation I'm pretty good..

5. For a biology experiment, the number of cells present is 1000. After 4 hours, the count is estimated to be 256000. What is the "doubling period" of the cells?
* How do you derive an equation from this? I tried and got x= 1/2 as my answer. Am i anywhere close?

6. A certain bacteria strain divides every 0.5 hours. If 500 bacteria are present in a culture, how many will be there
a) After 3 hours?
B) After 6 hours?
C) After n hours?

10. A certain strain of yeast cell doubles every 20 minutes. If there were 350 initially, how many will there be in 3 hours?

Again guys, I could really use the help.
Thanks so much!
As it's exponential growth your equation will be of the form $\displaystyle N = Ae^{kt}$ where N is the number, A is a constant (which I will soon denote $\displaystyle N_0$), k is the growth constant (dimensions: 1/time) and t is time.

To find A we use the initial condition; that is how many are at t=0. It becomes clear that N=A and so we let $\displaystyle A=N_0$ and define $\displaystyle N_0$ as the number at the start.

In your case you want to find k so rearranging my main equation gives

$\displaystyle k = \frac{1}{t} \ln{(\frac{N_0}{N})} = \frac{1}{4}ln(\frac{256000}{1000})$

You could number crunch to find k but it can be simplified since $\displaystyle 256 = 2^8$ and a factor of 1000 cancels to get

$\displaystyle k = 2ln(2)$.

Now we have our constant we can find out a relationship between doubling time and the constant k.

From the first equation: $\displaystyle N = N_0e^{kt}$ at doubling time $\displaystyle N = 2N_0$ and so $\displaystyle t_{double} = \frac{ln(2)}{k}$

In your case $\displaystyle k = 2ln(2)$ so $\displaystyle t_{double} = \frac{ln(2)}{2ln(2)} = \frac{1}{2} hour$ which makes your answer almost correct, you just need a unit.

You can use the main equation $\displaystyle N = N_0e^{kt}$ for all these, just changing the values as necessary

3. Originally Posted by brittt--
I really don't enjoy word problems. Can you give me a hand with the following? Once it's in an equation I'm pretty good..

5. For a biology experiment, the number of cells present is 1000. After 4 hours, the count is estimated to be 256000. What is the "doubling period" of the cells?
* How do you derive an equation from this? I tried and got x= 1/2 as my answer. Am i anywhere close?
$\displaystyle A=Pe^{rt}$

A = the ending amount of cells
P = the beginning amount of cells
r = rate of growth
t = time

$\displaystyle 256000=1000e^{4r}$

$\displaystyle 256=e^{4r}$

$\displaystyle \ln 256=4r$

$\displaystyle \frac{\ln 256}{4}=r$

We won't approximate here in order to avoid rounding errors.

To find out when the sample doubles,

$\displaystyle 2000=1000e^{\frac{\ln 256}{4}t}$

$\displaystyle 2=e^{\frac{\ln 256}{4}t}$

$\displaystyle \ln 2=\frac{\ln 256}{4}t$

$\displaystyle 4 \ln 2=\ln 256t$

$\displaystyle t=\frac{4 \ln2}{\ln 256}$

$\displaystyle t=.5$