Exponential Growth and Decay Question: A Bacteria Culture Contains 100 Cells and Grows at a Rate Proportional to its Size A bacteria culture contains 100 cells and grows at a rate proportional to its size. After an hour the population has increased to 420. 

a) Find and expression for the number of bacteria after $t$ hours

$P(t)=P(0)e^{kt}=100$ 

b) Find the number of bacteria after 3 hours. 

$P(t)=P(3)e^{k3}=$ (confused as to how to set this up)

d) When will the population reach 10,000?
$P(t) = 100e^{kt}$

$P(1) = 100e^{k*1} = 420$

$k = ln(4.2)$

b) is simply P(3)

c) solve P(x) = 10000
You know that "grows at a rate proportional to its size". Therefore you can write this in an equation:

$\frac{dP(t)}{dt}=kP(t)$ with some coefficient $k&gt;0$. This simple differential equation has for solution (after separation of variables) $P(t)=Ce^{kt}$, where $C&gt;0$ is a constant, that we determine from the initial condition: at $t=0$ you have 100 cells $\Rightarrow P(0)=100 \Leftrightarrow C=100$. You find $k$ with the second condition "After an hour the population has increased to 420", i.e $P(1)=420\Leftrightarrow 420=100 \times e^{k.1}\Rightarrow k=ln(4.2)$. 

For b): after 3 hours you have $P(3)=100\times e^{k.3}=100 \times(e^{ln(4.2)})^3=100\times 4.2^3=7408.8$.

For c): We want to find $t$ such that $P(t)=10000\Leftrightarrow 100\times e^{kt}=10000\Leftrightarrow e^{kt}=100\Leftrightarrow 4.2^t=100$ or $t=log_{4.2}100=\frac{ln 100}{ln 4.2}\approx3.2$