Question 1198940


general exponential models {{{y = a*b^t}}},  where {{{t}}} is the time from the starting moment and {{{b}}} is some arbitrary base


if the count in a bacteria culture was {{{600}}} after{{{ 20}}} minutes,we have 

{{{600 = a*b^20}}}............eq.1........solve for {{{a}}}

{{{a=600/b^20 }}}..........eq.1a


if the count in a bacteria culture was {{{1600}}} after{{{ 30}}} minutes,we have 

{{{1600 = a*b^30}}}............eq.2, sustitute {{{a}}}


{{{1600 = (600/b^20 )*b^30}}}............simplify

{{{1600 = 600*b^10}}}

{{{1600/600=b^10}}}

{{{8/3=b^10}}}

{{{b=root(10,8/3)}}}

then


{{{a=600/b^20 }}}..........eq.1a

if {{{8/3=b^10}}}=>{{{(8/3)^2=b^20}}}=>{{{b^20=64/9}}}

{{{a=600/(64/9) }}}
{{{a=675/8}}}

general exponential models {{{y = (675/8)*(root(10,8/3))^t}}}

What was the initial size of the culture?

the initial population "{{{a}}}" was{{{675/8}}}


The doubling period equation is {{{ b^t = 2}}}
{{{root(10,8/3)^t = 2}}}
{{{log(root(10,8/3)^t) = log(2)}}}
{{{t*log(root(10,8/3)) = log(2)}}}
{{{t= log(2)/log(root(10,8/3)) }}}
{{{t= log(2)/log(root(10,8/3)) }}}

{{{t=7.06695min}}}



Find the population after {{{60}}} minutes.

{{{y = (675/8)*(root(10,8/3))^60}}}
{{{y = 819200/27}}}
{{{y=30340.74}}}


When will the population reach {{{13000}}}

{{{13000= (675/8)*(root(10,8/3))^t}}}

{{{t = (8 log(4160/27))/log(1000)}}}
{{{t=5.83min}}}