How to solve this integral problem

A single injection of a drug is administered to a patient. The amount Q in the body then decreases at a rate proportional to the amount present, and for this particular drug the rate is 3% per hour. Thus, dQ/dt = - 0.03Q with Q(0) = Q0 where t is time in hours. If the initial injection is 4 milliliters Q(0) = 4 about how many hours after the drug is given will there be 2 milliliters of the drug remaining in the body? (Round answer to the nearest tenth of an hour.)

dQ/dt = - 0.03Q
-dQ/ 0.03Q = dt

Integrate both sides
(-1/0.03) ln(Q) = t + C
ln(Q) = -0.03t -0.03C

Q(0)=4
t=0; Q=4
ln(4) = 0.03(0) - 0.03C
C = -ln(4) /0.03 = -46.2098

ln(Q) = -0.03t -0.03 (-46.2098)
ln(Q) = -0.03t +1.386294

About how many hours after the drug is given will there be 2 milliliters of the drug remaining in the body?
ln (2) = -0.03 t + 1.386294
t = (ln(2)-1.386294) /-0.03
=23.1
about 23 hours

dQ/dt = -.03Q dQ/Q=-.03dt by integrating, lnQ=.03t+c
so Q=e^(-.03t+c) =Ae^-.03t where A=e^c=a constant
when t =0 Q=4 giving 4=A so Q=4e^-.03t when Q=2
e^-.03t=0.5 ie -.03t lne=ln0.5 ie t=(ln0.5)/-=.03=23.1hours.