# Differential Equations: Salt Concentration and logs

1. Mar 12, 2013

### Northbysouth

1. The problem statement, all variables and given/known data

A tank initially contains 180 gallons of water in which 8grams of salt are dissolved.
Water containing 9 grams of salt per gallon enters the tank at the rate of 3 gallons
per minute, and
the well mixed solution leaves the tank at the rate of 1 gallon per minute. The equation for the
amount of salt in the tank for anytime t.

2. Relevant equations

3. The attempt at a solution

My confusion is with the integrating factor.

First off I know:

Q' = rate at which salt enters - rate at which salt leaves

Q' = (9 grams/gal)(3 gal/min) - Q/(180 + 2t)

Simplify to Differential equation format

Q' + Q/(180 + 2t) = 27

Find the integrating factor.

u = e∫1/(180 + 2t) dt = e1/2 ln(180 + 2t)

My question is how do you know to move the 1/2 into the power to give the integrating factor u as:

u = (180 + 2t)1/2

I know it's a logarithmic law that I can move the coefficient into the power, but how do I recognize that I need to do this here? What's wrong with leaving the half in front as:

u = 1/2(180 + 2t)

Guidance would be appreciated. Thank you.

2. Mar 12, 2013

### Staff: Mentor

In this equation: u = e∫1/(180 + 2t) dt = e1/2 ln(180 + 2t)

you can rewrite the exponent on e on the far right as ln(180 + 2t)1/2.

That gives you eln(180 + 2t)1/2, which is the same as (180 + 2t)1/2.

The reason for doing this is to get the exponential in this form -- eln(A) -- which is equal to A, as long as A > 0.

There is no rule of logarithms that lets you change e1/2 ln(180 + 2t) to (1/2)(180 + 2t).