
Related Rates
Can someone help me with the following question?
A tank contains 100 litres of brine whose concentration is 3grams/litre. Three litres of brine whose concentration is 2 grams/litre flow into the tank each minute and at the same time 3 litres of miture flow out each minute.
(a). Show that the quantity of salt, Q grams, in the tank at any time t is given by:
Q = 200 + 100e^0.03t
The problem is I get my constant as a negative log.
Thanks.

dQ/dt = Salt in Salt out
dQ/dt = 6  3Q/100 Q(0) = 300
Note the equillibrium solution is Q= 200 Since we start with 300 Q will always be greater than 200
dQ/dt = 6  .03 Q
1/.03ln6.03Q = t + C
ln6  .03Q = .03t + C
6.03Q < 0 so ln6  .03Q = ln(.03Q 6)
ln(.03Q 6) = .03*t + C
at t = 0 ln(3) = C
.03*Q  6 = 3*e^(.03t)
Q = 200 +100e^(.03*t)