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)