Exponential growth
Say we start with one cell, put it in minimal medium, where it and its daughter cells will grow and divide once every hour:
In minimal medium , E. coli divides typically in 60 min., or 1 generation= 60 min.
We can calculate how long it will take to get a billion cells from just one:
Let g = number of generations. 2 gens. > 4 cells, 3 gens. > 8 cells, or
N (no. of cells) =1 x 2^{g} (starting with one cell).
If we started with 100 cells, after 1 gen. we would have 200 and after 2 gens. 400,
3 gens., 800 etc.
More generally, starting with N_{o} cells: N = N_{o} x 2^{g}
Since we want to know how much time it will take: express generations in terms
of time.
If we let t_{D} = the generation time, or doubling time, then the number
of generations that have passed during the time interval t is just t/t_{D}: So g =
t/t_{D}.
So now N = N_{o} x 2^{t/tD}. One can thus see that growth is
exponential with respect to time.
Now we could solve this equation for t, since we know we want N to be 1 billion, N_{o} is 1, and t_{D} is 1 hr. Taking the logarithm base 2 of both sides:
log_{2} (N/N_{o}) = t/t_{D}, or t = t_{D}log_{2}(N/N_{o}) = 1 X log_{2} (1,000,000,000/1) = log_{2}(10^{9})
But suppose your calculator doesn't do log base 2. No problem, convert to log base 10 ("log") or natural log base e ("ln").
log_{2}X = logX/log2 = logX/0.3 and log_{2}X = ln_{e}X/ln_{e}2 = lnX/0.69 (Also: 2^{x} = 10^{xlog2} and 2^{x} = e^{xln2})
So log_{2}(N/N_{o}) = log(N/N_{o})/log2 = t/t_{D} or log(N/N_{o}) = (log2/t_{D})t = Kt, where K=log2/t_{D} or K=0.3/t_{D}.
Or back to the exponential form: N/N_{o} =10^{Kt} or: N =N_{o}10^{Kt}
Or, since most scientific calculators have natural log functions:
N = N_{o}e^{Kt}, where K = ln2/t_{D} = 0.69/t_{D}, another common form of the exponential growth equation.
We could also have approached this question of rates of change of N with time more naturally using calculus (Note: familiarity with calculus is not necessary for this course.) If you have a million cells, then after one generation time you’ll have gained 1 million. If you had 100, you would’ve gained 100. In general, the rate of increase of N with time is just proportional to the number of cells you have at any moment in time, or: dN/dt = KN
Separating variables: dN/N = Kdt.
Integrating between time zero when N = N_{o} and time t, when N = N:
lnN  ln N_{o} = Kt  0, or ln(N/N_{o}) = Kt, or N = N_{o}e^{Kt}
We can calculate the constant K by considering the time interval over which N_{o}
has doubled. This time is the doubling time, t_{D}. For that condition:
N/N_{o} = 2 = e^{KtD}. Taking the natural logarithm of both sides:
ln2=Kt_{D}, or K=ln2/t_{D}, exactly as above.
In summary:

Base 2  Base 10  Base e 
Exponential form  N = N_{o}2^{K2t}  N = N_{o}10^{K10t}  N = N_{o}e^{Ket} 
Logarithmic form 
log_{2}(N/N_{o}) = K_{2}t 
log(N/N_{o}) = K_{10}t  ln(N/N_{o}) = K_{e}t 
Definition of constant 
K_{2} = 1/t_{D} 
K_{10} = log_{10}2/t_{D} = 0.3/t_{D} 
K_{e} = ln2/t_{D} = 0.69/t_{D} 
All this looks worse than it is. Exponential growth using a base of 2 is intuitively obvious. And once you see the derivation, the exponential growth equation using log or ln can be simply applied to problems using a calculator. You just have to keep track of what you know and what you are after.
Graphically, the depiction of exponential growth looks like this: 
Or, with the ordinate (Yaxis) plotted on a logarithmic scale, a semilog plot: 


In reality, there's a lag before cells get going, and there's a limit (thankfully) to cell density, as nutrients become exhausted and/or toxic excretions accumulate. The final plateau is called stationary phase:
© Copyright 2000 Lawrence Chasin and Deborah Mowshowitz Department of Biological Sciences Columbia University New York, NY