please solve An initial population of bacteria of 100 double

please solve

An initial population of bacteria of 100 doubles every day. Each bacterium lives at least 10 days. The size of the population of bacteria after 7 days is 100 bacteria. 12800 bacteria. 10 bacteria. 700 bacteria.

Solution

Solution:

Y= a (1 + r)^x

a = the number of bacteria there are initially = 100
r = the rate of increase which is 1.00 (ie 100% or \"double\")
x = 7 (ie the number of time periods)

Therefore Y = 100(1+1)^7 = 100(2)^7 = 100 x 128 = 12800 bacteria