Give a doubling time for an exponentially growing quantity.?

Question

The number of cells in a tumor doubles every 6 months. If the tumor begins with a single cell, how many cells will there be after 3 years? after 6 years?

Answer 1

The exponential formula for growth would be

n(t) = n(0)* e^(t/tc), where n0 is the initial number of cells, and tc the "time constant" for growth. This can also be written

n(t)/n(0) = e^(t/tc)

You are given that when t = 6mo, n(6)/n(0) = 2

therefore 2 = e^(6/tc); solve this for tc:

ln(2) = 6/tc, and tc = 6/ln(2) = 8.656

The equation is now n(t)/n(0) = e^t*/8.656

where t is in months. For t = 3 years, that is 36 months, so n(36)/n(0) = e^(36/8.656) = 64.0 For t = 6 years, or 72 months n(72) = e^(72/8.656) = 4.10*10^3