Question 862226
Your model is ... not bad.  Try this:  {{{900=300e^(k*6)}}}.

Simplify and solve for k.
{{{3=e^(k*6)}}}
{{{ln(3)=ln(e^(6k))}}}
{{{ln(3)=6k*1}}}
{{{highlight_green(k=(1/6)ln(3))}}}
{{{highlight(k=0.183)}}}


Now, let t=12 hours, and find the bacteria count.
{{{C=300*e^(0.183*12)}}}
{{{C=300*9}}}
{{{highlight(C=2700)}}}.


Notice that my solution simply used three significant figures while yours used two significant figures in the k value.  That is why you found C=2601.