Question 771049
{{{A(t)=6000e^(-.05t)}}}
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(A)Find the amount present after 12 days
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This means, find A when t=12, or find {{{A(12)}}}.

{{{A(12)=6000*e^(-0.05*12)}}}
{{{A(12)=6000*0.5488}}}
{{{A(12)=3300}}}, not being sure how much accuracy your given values really allow.



(B)Find the half-life of the material
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When A(t) is 3000.
{{{3000=6000*e^(-0.05t)}}}, and you want to find this t.
{{{1/2=1*e^(-0.05t)}}}
{{{ln(1/2)=ln(e^(-0.05t))}}}
{{{ln(1/2)=-0.05t*ln(e)=-.05t}}}
{{{t=(-1/0.05)*ln(1/2)}}}
{{{t=-20*ln(1/2)=-20*(-0.693)}}}
{{{t=13.9}}} years, or, not knowing how much accuracy is given, maybe 14 years is half life.