If p(t)=200, then 200=250-120*2.8^-0.5t. So 120*2.8^-0.5t=50, 2.8^-0.5t=50/120 and 2.8^0.5t=120/50=2.4.
We need to take logs of both sides: 0.5t*log(2.8)=log2.4, so t=log2.4/0.5*log2.8=2log2.4/log2.8=1.7006 months.
Let's put this value of t into the original equation: p(1.7006)=250-120*2.8^-.5*1.7006=200. Please note that because of the logs t=1.7006 is inevitably an approximation. However, because the number is so close to 1.7 months, there may in fact be approximations in other values in the function, so the solution 1.7 months may be the exact answer.