Question 1077241
From 6 years ago to now:

{{{950=1710*e^(kt)}}}, k will be found negative in value.

{{{e^(kt)=950/1710}}}

{{{e^(kt)=95/171}}}

{{{e^(kt)=5/9}}}

{{{ln(e^(kt))=ln(5/9)}}}

{{{kt=ln(5/9)}}}, and t quantity used was t=6;

{{{k=(1/6)ln(5/9)}}}

{{{k=-0.09796}}}

MODEL:  {{{highlight_green(A=A[o]e^(-0.09796t))}}}


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If {{{A=150}}} and {{{A[o]=950}}}, what is t?
You do this on your own?