A type of algae grows continuously such that its population doubles after 3 days. What's the population after 10 days? 
A type of algae grows continuously so that its population doubles in 3
  days. Given a beginning population of 100 algae cells per milliliter
  of water, to the nearest whole number, how many algae cells would you
  expect at the end of 10 days?

I start off by dividing 10 by 3 giving me 3 doubles in population:
$$100\rightarrow200\rightarrow400\rightarrow800$$
Now I have to deal with the extra day. I get $800+\frac{1}{3}\cdot800\approx1067$ cells per milliliter. This is incorrect. Where is my logic or math wrong? How should I solve this problem?
Thanks! Your help is appreciated!
Max0815
 A: You are right that after $9$ days you've had $3$ doublings. You have one day, or one third of a doubling period to worry about. Your linear answer is wrong. This is a problem in exponential growth. The growth factor is the same for all time intervals of the same length. You know that factor is $2$ for $3$ day intervals so it is $z = \sqrt[3]{2}$, the cube root of $2$, for a one day interval. So to answer your question, find the cube root $z$ of $2$ accurately enough to know $800z$ to the nearest whole number.
If you know about logarithms you can work with a formula. The number $A$ of organisms at time $t$ days starting from $100$ at $t=0$ is
$$
A = 100 \times 2^{t/3}.
$$
Substitute $t = 10$ and solve the equation for $A$, then round to the nearest integer.
You can tell that will give the same as the first method since
$$
2^{10/3} = 2^3 2^{1/3} = 8 \sqrt[3]{2}.
$$
A: As John douma has mentioned in the comments you have to use the continuous model: $P'(t)=k\cdot P(t)$, where $P(t)$ is the population of the algae cells.
First of all we solve this differential equation by the method of Separation of variables.
$P'(t)=k\cdot P(t)$
$\frac{dP}{dt}=k\cdot P(t)$
Dividing the equation by $P(t)$ and multiplying the equation by $dt$.
$\frac{1}{P(t)} \ dP=k \ dt$
Integrating both sides
$\int \frac{1}{P(t)} \ dP=\int k \ dt$
$\ln(P)=k\cdot t+c$
To obtain $P$ on the LHS we take both sides as an exponent of the number $e$.
$e^{\ln(P)}=e^{k\cdot t+c}$
$P=e^{k\cdot t}\cdot e^c$
Replacing $e^c$ by $C$
$P=C\cdot e^{k\cdot t}\quad (*)$
To determine the constants C and k we use the following information:

  
*
  
*Given a beginning population of $100$ algae cells per milliliter of water ...
  

The equation is $P(0)=100\Rightarrow P(0)=C\cdot e^{k\cdot 0}=C\cdot 1=100\Rightarrow C=100$


  
*... so that its population doubles in 3 days
  

That means after 3 days the population is $200$: $P(3)=200$
$P(3)=100\cdot e^{k\cdot 3}=200$
$e^{k\cdot 3}=\frac12$
Taking $\ln()$ on both sides.
$3\cdot k=\ln\left( \frac12\right)$
$k=\frac{\ln\left( \frac12\right)}{3}$
Now we can use (*) to obtain $P(t)$
$P(t)=100\cdot e^{\ln(2)\cdot \frac{t}3}=100\cdot \left(e^{\ln(2)}\right)^{ \frac{t}3}=100\cdot 2^{\frac{t}{3}}$
