Question 63050
amount of a radioactive tracer remaining after t days is given by A=Aoe^-0.058t, where Ao is the starting amount at the beginning of the time period. how many days will it take for one half of the original amount of decay?
A=(1/2)Ao
{{{(1/2)Ao=Aoe^(-0.058t)}}}
{{{(1/2)Ao/Ao=Aoe^(-0.058t)/Ao}}}
{{{1/2=e^(-0.058t)}}}
{{{ln(1/2)=ln(e^(-0.058t))}}}
{{{ln(1/2)=-0.058tln(e)}}}  ln(e)=1
{{{ln(1/2)=-0.058t(1)}}}
{{{ln(1/2)/-0.058=-0.058t/-0.058}}}
{{{-ln(1/2)/0.058=t}}}
{{{t=11.95081346}}}
About 12 days.
Happy Calculating!!!