This is how I would solve it (essentially as you did):
Using the change of base formula, we may write:
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
thats about 10000
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
t = 45.15 days <==ANSWER
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
Hello, Nickod777!
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: . days. . You are correct!
The population function is: . .where is measured in days.
We have: .
Take logs (base 10): .
. . . . . . . . . . . . . . . .
n . . . . . . . . . . . . . . . . . . . . .