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.
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/.03ln|6-.03Q| = t + C
ln|6 - .03Q| = -.03t + C
6-.03Q < 0 so ln|6 - .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)