Question

A certain bacteria has an exponential growth pattern. after 3 hours there are 51 bacteria, and after 12 hours there are 685 bacteria. how many bacteria will be present after 24 hours?

Answer 1

when t = 3.. x =\(x_o\) = 51
then when t = 12...x = 685

therefore
\[\ln(\frac{x}{x_o} )= kt\]
\[\ln (\frac{685}{51} = k(12 -3)\]
\[\ln (13.4313) = k(9)\]
\[\frac{\ln (13.4313)}{9} = k\]
\[k = 0.2886\]
now use
\[x = x_oe^kt\]
\[x = 51 e^{(0.2886)(24-3)})\]
\[x = 51e^{6.0606}\]
\[x = 51(428.6325)\]
\[x = 21,860.25\]
\[x = 21,860\]
got it?