Question 880711
{{{H(t)=100/(1+150e^(-0.32t))}}}
{{{100/(1+150e^(-0.32t))=50}}}
{{{1/(1+150e^(-0.32t))=1/2}}}
{{{1+150e^(-0.32t)=2}}}
{{{150e^(-0.32t)=1}}}
{{{e^(-0.32t)=1/150}}}
{{{-0.32t=ln(1/150)}}}
{{{t=-ln(1/150)/0.32}}}
or approximately,
{{{t=15.66}}}
Almost 16 years old.