Differential equation for bacterial growth I am assigned with a question which states the rate of a microbial growth is exponential at a rate of (15/100) per hour. where y(0)=500, how many will there be in 15 hours?
I know this question is generally modelled as: 
$y=y_0*e^{kt}$
However, my solution ended up being modelled as :
$y=e^{kt}*e^{c}$  via $y'=ky$
The resulting equation was:
$y=e^{kt}*e^{ln500}$
I ended up getting $y(15)=4743.86$ which is the same answer for both methods.
I'm wondering how the general equation was modelled, and if someone could explain how I could tidy up my equations.
Thanks. 
 A: The general equation can be modeled in the following way:
If we call the bacteria $B$, and time (obviously) $t$, then $B$ is clearly a function of time $B(t)$. Notice that $B$ is proportional with the rate of change.
$$\frac{dB}{dt} \propto B
\quad\implies\quad \frac{dB}{dt}  = kB,$$
Where $k$ is the proportionality constant. As we can see, this is just a separable differential equation, and to solve this we separate the variables and integrate in the following way:
$$\left(\frac{dB}{dt}\right)\frac{1}{B} \,=\, (kB)\frac{1}{B} \;\implies\; \frac{1}{B}\frac{dB}{dt}dt \,=\, kdt$$ $$\implies\; \int \frac{1}{B}dB \,=\, \int k\,dt \;\implies\; \ln|B|+C_1 \,=\, k(t+C_2) $$
$$\implies\; \ln|B| \,=\, kt+(kC_2-C_1),\quad \text{and letting}\quad kC_2-C_1=C_3,$$ $$\implies\; \ln|B|=kt+C_3.$$ $$\because B\geq 0\quad\quad (\text{you can't  have negative bacteria}) $$ $$\therefore \ln(B)=kt+C_3 
\;\implies\; e^{\ln(B)}=e^{kt+C_3} $$ $$\implies B=e^{kt}e^{C_3},\quad \text{and letting}\quad e^{C_3}=C,$$ $$\therefore B=Ce^{kt} \quad\quad (\text{exponential growth})$$
B(t) plot:

Note that $C$, and $k$ are determined by the initial conditions. I'm guessing there can be some confusion about how $e^{C_3}$ can be equated to $C$, but remember that $e$, $C_1$, $C_2$, $C_3$, and $C$ are all constants, therefore, it wouldn't make a difference if we equate $e^{C_3}$ with $C$.
I hope i did answer your question.
A: $e^{\ln 500}$ is equal to 500. Notice that if $y = e^x$, taking the natural log of both sides gives you $\ln y = \ln (e^x) = x$. Thus, to undo this operation, take each side as the power of $e$ to get $e^{\ln y} = e^x$, which must be $y = e^x$. 
