Tuesday, May 17, 2011


Today in class, we learned about logarithms. We first went over how to find the "log" button on a calculator. You press CTRL 10^x. We then started on the 9.5 notes and sketched a graph of the exponential function y=10^x. From there we found the inverse by switching x and y. This is where the new function, logarithms, comes in. We found that the inverse of y=10^x is x=10^y, or the log formula. After graphing both we compared and contrasted the two equations.
y=10^x y=log x or x=10^y
Domain: all reals x>0
Range: y>0 all reals
x-int.: none (1,0)
y-int.: (0,1) none
Asymptote: x-axis y-axis

After, we did some examples without our calculators. The first problem we looked at was log 10^100. In order to figure it out, we set in up as 10^? = 100. From there it was easy to figure out that the answer was two. We then moved on to decimals. The next problem was log 10^0.1. In order to figure it out we changed it to 10^? = 1/10. We then figured out that our answer was -1.
Some more examples are:
log 10^ 1,000 = 3 log 10^ 10 = 1 log 10^ 1000000 = 6
log 10^ 5 = .669 log 10^ 30 = 1.477

It is important to notice that log 10^ (-100) does not work. This is because you can't raise 10 to a value and get a negative number. Another important thing to notice is that the base does not have to be 10. A problem can be log 9^ 81.

The next set of problems that we did were very similar to the previous examples. However, they did not included a base number. When logs do not have a base number, it can be assumed that 10 is the number. "Log x is called the common logarithm, so log 10^x is the same as log x."
log 10000= 4 -This is because you assume the equation is written as log 10^ 10000.
log (o.oo1) = 3 log 10 = 1 log 7 = .8451 log 316 = 2.4997

We were then introduced to another set of equations. Log x = 4. It looked difficult, however, using the two methods we previously learned, it was pretty simple. First we put in 10 as the base because we learned that you can assume a log without a base is 10. Then we changed the equation around to read 10^4 = ?. This equation resembled the ones we had done previously. We then used our green cards and found that x = 10,000.

The last lesson of the day was the conceptual question., "Between what two integers is log 7598?" To solve this we asked ourselves 10^ ? = 7598. We then plugged in number for 10^ ? and found minimum and maximum values of 3 and 4. These two numbers worked because 10^ 3 = 1,000 and 10^ 4 = 10,000. We knew that 3 and 4 were the two integers because 7598 fit in the middle of them.

log 172: log .4:

10^ ? = 172 10^ ? = .4

10^ 2 = 100 10^ 0 = 1
10^ 3 = 1,000 10^ -1 = .1

Answer is 2 and 3 because 172 fits in Answer is 0 and -1 because .4 fits in
between 10^2 and 10^3. between 10^0 and 10^-1.

If you still need help the 9.5 notes are located on moodle. Also the homework for tonight is 905 - 1, 3-11, 14-16, 18, 21-23.

Wednesday, May 11, 2011

Chap 9 Section 3: Continuous Compounding

Hello class. At the beginning of the day, Mr. Cope gave a 5 minute review on sections 9.1 and 9.2. He simply taught exponential growth (9.1) and exponential decay (9.2). He taught the equation y=ab^x. A being the number you begin with, b being the growth factor (growth b>1, decay 0<b<1) and the x being the number of years being used. This was just a quick review of what was taught on Monday and Tuesday.

Next Mr. Cope handed back the tests and homework quizzes we took last week. Unfortunately, Teacher Logic no longer shows the class average so there will be no comparison to period 1 in this blog post. Anyway, we were able to compare our answers to Mr. Copes key and then we moved on to today's lesson.

Today we learned about continuous compounding. We began with a chart showing a bank pays 100% interest for 1 year. So we went through annually, quarterly, monthly, daily, and hourly using the equation P(1+r/n)^nt. The final value (hourly) was very close to e (2.71828) or Euters number which is used to represent exponential growth. So the formal equation for Continuously Compounded Interest is a=Pe^rt. P stands for the original number, e stands for Euters number, r stands for either the growth of decay (growth=if number is positive, decay if number is negative), and t stands for the number of years. We then did an example 1 which was "Suppose $1800 is invested at an annual interest rate of 7% compounded continuously and the money is what happens in 5 years. So the equation would be 1800e^.07(5). You then would solve it on the calculator and your answer should be 2554.32.

Moodle notes: http://gbs-moodle.glenbrook225.org/moodle/mod/resource/view.php?id=13797

Friday, May 6, 2011


So I saw this pic whilst browsing the internet and I was like, BAM! MR. COPE!