So, you have N(t)=Noe^t/2 We are looking for the time when the # of bacteria doubles, so we are solving for t. We set it up as: 2(No)=Noe^t/2 We can now get rid of the No, making the equation much for solvable. 2=e^t/2. Now we take the natural log (ln) of both sides because this will eliminate the e, because loge^e is equal to 1. We get: ln2=t/2 multiply by 2 on both sides to cancel out the 2 and get t alone. We get: 2ln2 = t or approximately: 1.4 units (whatever your units were)
2 = 1 * e^(t/2) ln(2) = ln[e^(t/2)] ln(2) = t/2 t = 2ln(2)
