i need help with algebra

Question

A certain radioactive isotope has leaked into a small stream. Two hundred days after the​ leak, 33​% of the original amount of the substance remained. Determine the​ half-life of this radioactive isotope.

Answer 1

0.33 of the material is left after t = 200. figuring on a continuous decay

So we can write 0.33A = Ae^200r. where r is the rate of decay.

then 0.33 = e^200r

using logs to bring down the exponent, we get

ln(0.33) = ln(e^200r) = 200r(ln e) = 200r. since ln(e) = 1

then. ln(0.33)/200 = -0.0055433 about

then .50A = Ae^-0.0055433t

Using the same process. 0.50 = e^-0.0055433t ------> ln(50) = -0.0055433t ---> 125 days