# Pre-calc amoeba doubling every 4 days

• Feb 3rd 2013, 12:30 PM
Nickod777
Pre-calc amoeba doubling every 4 days
A single-celled amoeba doubles every 4 days. How long would it take one amoeba to produce a population of about 10,000 amoebae?

I have this problem up above....

I have come up with my own formula, so I posted it on my facebook, then all of a sudden I got 20 or so replies on different formulas. Can you guys help me decide which could be right? I really feel like my formula is right.

1. My formula 4 ( ln 10000 / ln 2) = 13.2877*4= 53.15 days....
I used this as proof on wolfram alpha. 2^x = 10000, and it came to 13.2877 or so, then I multiplied by 4.

2. A friend says this
4 ( ln 10000 / ln 2) + 4 = 57.15084 days which he used this as proof....
(2^(57.15084 / 4)) / 2 = 9 999.98351

3. Another friend says this

10,000 = 4(2)^(t/4)
2500 = 2^(t/4)
log 2,500 = (t/4) log 2
4 log 2,500 / log 2 = t

4. Then finally this guy shows up out of nowhere and does this....
P = Po*e^(k*t)
P=pop after time
Po= original population
k= rate constant
t=time
2Po = Po*e^(k*4)
2 = e^(k*4)
ln(2) = k*4
k = (1/4)ln(2)
k = 0.173

10000 = (1)e^(0.173*t)
ln(10000) = 0.173*t
t = 53.2 days
• Feb 3rd 2013, 12:53 PM
MarkFL
Re: Pre-calc amoeba doubling every 4 days
This is how I would solve it (essentially as you did):

$\displaystyle P(t)=2^{\frac{t}{4}}=10000\,\therefore\,\frac{t}{4 }=\log_2(10^4)=4\log_2(10)\,\therefore\,t=16\log_2 (10)$

Using the change of base formula, we may write:

$\displaystyle t=\frac{16\ln(10)}{\ln(2)}\approx53.15$
• Feb 3rd 2013, 12:58 PM
Nickod777
Re: Pre-calc amoeba doubling every 4 days
Awesome, thanks for the quick reply.
• Feb 3rd 2013, 03:17 PM
Soroban
Re: Pre-calc amoeba doubling every 4 days
Hello, Nickod777!

Quote:

A single-celled amoeba doubles every 4 days. How long would it take
one amoeba to produce a population of about 10,000 amoebae?

My formula: .$\displaystyle 4\frac{\ln 10,\!000}{\ln 2} \:\approx\:53.15$ days. . You are correct!

The population function is: .$\displaystyle P(t) \:=\:2^{\frac{1}{4}t}$ .where $\displaystyle t$ is measured in days.

We have: .$\displaystyle 2^{\frac{1}{4}t} \:=\:10,\!000$

Take logs (base 10): .$\displaystyle \log(2^{\frac{1}{4}t}) \:=\:\log(10,\!000)$

. . . . . . . . . . . . . . . . $\displaystyle \tfrac{1}{4}t\log(2) \:=\:4$

n . . . . . . . . . . . . . . . . . . . . . $\displaystyle t \:=\:\frac{16}{\log(2)} \:=\:53.15084952$