Question 364239
{{{N(t)=N[0]e^(kt)}}}
{{{0.8=2.4e^(k(40))}}}
{{{e^(40k)=1/3}}}
{{{40k=ln(1/3)}}}
{{{k=(1/40)*ln(1/3)}}} or approximately,
{{{k=-0.02747}}}
.
.
.
{{{(1/2)N[0]=N[0]e^(kt[H])}}}
{{{kt[H]=ln(1/2)}}}
{{{t[H]=ln(1/2)/(1/40)ln(1/3)}}}
{{{t[H]=40(ln(1/2)/ln(1/3))}}} or approximately,
{{{highlight(t[H]=25.2)}}}hrs