Block Days, Wed & Thurs 4/30 & 5/1 Solving equations using natural log and log continued. Newton's Law of Cooling Primer

Post date: May 1, 2014 8:42:08 PM

Notes:

In the last couple days we learned how to solve log equation algebraically, now let's look at how to solve log equations graphically.

Example 1: Solve graphically:

Method 1:

The window must be adjusted to show a sufficient amount of the x-axis to locate the intersection point (2nd CALC, #5). The answer is x = 32.

Method 2:

This problem can also be solved by setting the equation equal to 0 and finding the x-intercepts, or zeros (2nd CALC, #2), of the function.

Now let's apply this to real life situations, we'll be using natural log and Newton's Law of Cooling. According to Newton’s law of cooling equation, the change in the temperature of an object will depend on the difference between its own temperature and the temperature of its surrounding environment. We can give the statement in mathmeatical form as below:

\[ T(t)= T_a + (T_o-T_a) e^{-kt} \]

Example 2: As part of his summer job at a resturant, Jim learned to cook up a big pot of soup late at night, just before closing time, so that there would be plenty of soup to feed customers the next day. He also found out that, while regrigeration was essential to preserve the soup overnight, the soup was too hot to be put directly into the fridge when it was ready. He notice that the equation

\[ T(t)= 5 + 95 e^{-kt} \]

gives the tempeature of the soup after it begins cooling from 100 degrees. How long does it takes for the soup to be cool enough to put into the regrigerator.(T=20 degrees)?

$ T(t)=20 $

We need to wait until , so at that time:

\[ 20= 5 + 95 e^{-0.054t} \]
$ t $

This equation can be solved for in much the same way as before. Subtracting 5 from both sides and dividing by 95 we get:

\[ \frac{15}{95} = e^{-0.054t}~~~~~\frac{95}{15}= 6.333 = e^{0.054 t} \]

Taking logarithms of both sides, we find that

$ \ln(6.33)= 1.84 $
\[ \ln(6.33) = \ln(e^{0.054t}) = 0.054 t \]

Thus, using the fact that we have

\[ t= \frac{1.84}{0.054}=34.18 \]

Thus, it will take a little over half an hour for Jim's soup to cool off enough to be put into the refrigerator.

Other Resources:

Natural Log Properties and Using Natural Logs for solving equations

HW: Assign #6 WS pg 21 (1-19 odds no 11)