Question 825694
The exponential decay model is based on {{{A=I*e^(kt)}}}; some people might prefer a {{{-k}}}, but however the choice, consistency is needed.


A = amount at time t
I = initial amount, for t=0
k = a constant
t = time in years


First, find k.
{{{A/I=e^(kt)}}}
{{{ln(A/I)=ln(e^(kt))}}}
{{{ln(A/I)=kt*1}}}
{{{k=(1/t)ln(A/I)}}}
'
The given halflife is for t=7340 years, and for which {{{A=(1/2)I}}};
{{{k=(1/7340)ln(1/2)}}}
{{{k=-9.44*10^(-5)}}}
'
Not sure if this will render properly but the model is:
{{{highlight(A=I*e^(-0.0000944*t))}}}


Your specific question will use A=0.20, and I=1.  You would then solve the model equation for t.
Recall from, "First, find k", a slight change in step gives:  {{{highlight(t=(1/k)ln(A/I))}}}; just substitute the values known.