# Math Help - inhibited growth

1. ## inhibited growth

There is a problem in my textbook that is exactly like one of the examples, but just uses different numbers.

Problem: A tank initially contains 400 gal of brine in which 100 lbs of salt are dissolved. Brine containing 1/10 lb of salt per gallon is run into the tank at 20 gal/min, the mixture being drained off at the same rate. How many lbs of salt remain in the tank after 30 minutes?

In the example, we separate variables and integrate. I use the same method to solve this problem, it just has different numbers and I get: 40-60e^(-t/20). The books's answer is 40 + 60e^(-t/20).

What am I doing wrong?

2. I got it.

When going through the problem you deal with the quantity ln(40-x). But 40-x would be negative at t=0, so we must be dealing with -(40-x)=(x-40)

3. Looks to me like you probably integrated

$\dfrac{1}{40-x}$ and got $\ln|40-x|,$ whereas you should have gotten $-\ln|40-x|.$ You have to do a $u$ substitution in order to use that rule!