The class began as any class would with Mr. K. putting the finishing touches on the day's lesson, while students assembled to talk about their various "Spring Break adventures". Then Mr. K. started with a very important notice, which I will repeat:
FLICKR ASSIGNMENT IS DUE BY MIDNIGHT TONIGHT!!!
Then, he continued to remind us about the DEVELOPING EXPERT VOICES project. It went as follows:
1. Make sure everyone has their due date set for the publication of the project.
2. E-mail Mr. K. so that he can invite you to the new DEV Blog. This is where your project is to be published, so join that ASAP, especially if you're project is due soon.
3. Finally both tomorrow and Wednesday's class will be work periods for any homework (Exercise 21) as well as the DEV projects.
Finally, we got down to some math, yay :D
We began with some review from before the break. He went over both the multiplication and division of powers. The main thing to remember is that the bases must be the same to be able to simplify the expression. In the case of multiplication, the exponents can be added. For the opposite of multiplication (division), you would do the opposite of addition (subtraction). From these examples, we can conclude:
Next, Mr. K. applied this to logarithms. ( Which, I will remind you, are EXPONENTS!).
This led to two of the LAWS OF LOGARITHMS:
Mr. K. followed this up with the presentation of a couple questions that dealt with the combination of both multiplictaion and division. They can be seen HERE.(Slide 6/8)
THEN, Mr. K. handed back our trigonometric identities test! This was somewhat of a distraction to many people. There were groans from a lot of the class (including myself) as we flipped through the papers to see where we had made the mistakes.
When all the excitement had died down, Mr. K. was showing us the power within a power. In these examples, the base of the power is a power itself. See there--->. When dealing with such problems, the inside exponent(s) is(are) multiplied by the outside exponent.
However, when this is applied to Logarithms it becomes a little more complicated. There is the usual base, but the argument also contains a base as well as an exponent. If the bases are the same, then obviously the logarithm is equal to the the exponent in the argument. But if they are different, there are two ways to solve it. First (<---), create a 'power within a power' situation out of the argument with the same base as that of the logarithm. For example, in the image to the right the base is 2. The base in the argument is 8 so it can be simplified to 2^3. This makes the bases equal. Then just multiply the inside and outside exponents, as explained above, and you get 2^15. Now that the bases are the same, follow the "rule" above (and seen below and to the left) and get 15 as the final answer.
The second way of solving is by rearranging the expression as seen here-->
This can be rearranged because "log base 2 of 8" is raised to the exponent 5. Since logarithms are exponents you can multiply the exponent 5 by the exponent (log base 2 of 8) to get what is seen to the right. Then it can be determined that "log base 2 of 8 is equal to 3. 3 multiplied by 5 is 15 which gives the correct answer.
Mr. K. then finished by generalizing that last bit as a RULE:
and then gave us a couple of examples seen HERE. (Slide 8/8).
Well, that wraps up my scribe post for the night.
REMEMBER: FINISH YOUR FLICKR ASSIGNMENTS BY MIDNIGHT!!!
Oh yeah, and the next Scribe is:
THEN, Mr. K. handed back our trigonometric identities test! This was somewhat of a distraction to many people. There were groans from a lot of the class (including myself) as we flipped through the papers to see where we had made the mistakes.
When all the excitement had died down, Mr. K. was showing us the power within a power. In these examples, the base of the power is a power itself. See there--->. When dealing with such problems, the inside exponent(s) is(are) multiplied by the outside exponent.
However, when this is applied to Logarithms it becomes a little more complicated. There is the usual base, but the argument also contains a base as well as an exponent. If the bases are the same, then obviously the logarithm is equal to the the exponent in the argument. But if they are different, there are two ways to solve it. First (<---), create a 'power within a power' situation out of the argument with the same base as that of the logarithm. For example, in the image to the right the base is 2. The base in the argument is 8 so it can be simplified to 2^3. This makes the bases equal. Then just multiply the inside and outside exponents, as explained above, and you get 2^15. Now that the bases are the same, follow the "rule" above (and seen below and to the left) and get 15 as the final answer.
The second way of solving is by rearranging the expression as seen here-->
This can be rearranged because "log base 2 of 8" is raised to the exponent 5. Since logarithms are exponents you can multiply the exponent 5 by the exponent (log base 2 of 8) to get what is seen to the right. Then it can be determined that "log base 2 of 8 is equal to 3. 3 multiplied by 5 is 15 which gives the correct answer.
Mr. K. then finished by generalizing that last bit as a RULE:
and then gave us a couple of examples seen HERE. (Slide 8/8).
Well, that wraps up my scribe post for the night.
REMEMBER: FINISH YOUR FLICKR ASSIGNMENTS BY MIDNIGHT!!!
Oh yeah, and the next Scribe is:
2 comments:
Hi Craig,
Happy Birthday!!!
Thanks for a scribe that made me feel like I was right in class with you!
Best,
Lani
HBD Craig, I nominate you for the hall o' fame. Finally got a chance to read it and it was fantastic. But not as good as 300. Nice try though. Still HoF worthy.
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