Question 1158943
{{{6.5=13e^(k*24)}}}, using continuous exponential decay model

{{{13e^(24k)=6.5}}}

{{{e^(24k)=6.5/13}}}

{{{ln(e^(24k))=ln(6.5/13)}}}

{{{24k=ln(6.5/13)=ln(1/2)}}}

{{{k=ln(1/2)/24}}}

{{{k=-0.02888}}}


If 2 milligrams, how much time, x?
{{{ln(e^(-0.02888x))=ln(2/13)}}}

{{{-0.02888x=ln(2/13)}}}

{{{x=-ln(2/13)/0.02888}}}

{{{highlight(x=65)}}}{{{minutes}}}