Question 1127637
A=Be^-t/2


You need to write it properly.  Not sure if you mean as  {{{A=Be^(-t/2)}}}.


Using natural log base,


{{{ln(A)=ln(Be^(-t/2))}}}


{{{ln(A)=ln(B)+ln(e^(-t/2))}}}


{{{ln(A)-ln(B)=-(t/2)*ln(e)}}}


{{{ln(A)-ln(B)=-t/2}}}


{{{t/2=ln(B)-ln(A)}}}



{{{t=(1/2)(ln(B)-ln(A))}}}