Initial Amount: Ao=30 grams
k=0.1253
Amount after t days: A(t)
A(t)=Ao*e^(-kt)
Half-life is the time t=T when A(t)=Ao/2:
Ao/2=Ao*e^(-0.1253T)
Dividing both side by Ao:
1/2=e^(-0.1253T)
Applying natural logarithm (ln) to both sides of the equation:
ln(1/2)=ln[e^(-0.1253T)]
ln1-ln2=-0.1253T*lne
0-ln2=-0.1253T*1
-ln2=-0.1253T
Dividing both sides of the equation by -0.1253:
-ln2/(-0.1253)=T
T=ln2/0.1253
T=0.693147181/0.1253
T=5.531900882
T=5.5 days
The substance's half life is 5.5 days.
