Friday, April 6, 2007

SsSsSs-Spring-cribe


AsianTown.NET



MORNING CLASS

--- Hey guys, I’m Thursday’s scribe and I am obliged to reiterate and discuss what happened in today’s classes. Before we dove into Logarithms, Mr. K. had some announcements. One of the announcements pertain to the FLICKR assignment. All pictures’ link must be posted up on the blog and the evaluation sheet regarding the assignment should also be submitted. He pointed out the random pictures that some people posted, which were all related to the particular theme --- 300. After the FLICKR assignment, Mr. K. discussed the Developing Expert Voices Project. He showed us several examples of different presentations of a project. One was when the student incorporated SIMS into his project. One sang the project which I thought was cool. I was going to do something like that but not for the entire project. MAYBE! HAHA! They certainly inspired the class and further increased the level of the bar that we have to overcome. Definitely, we are pressured, but I am excited about DEV. We also learned today that our blog is viewed by different people all over the world.
Now, with the real stuff --- LOGARITHMS! Let’s start by remembering what a logarithm is.

LOGARITHM IS AN EXPONENT!!


--- Then, say that five more times! Drill it in your head so that you will do well in this unit. Mr. K. displayed a few questions as warm-up questions that concern the Laws of Logarithms we learned before the Spring Break:



logbasePOWER = Exponent

logb MN = logb M + logb N

logb (M/N) = logb M - logb N

We were asked to simplify the following logarithms:



* It can also be answered it like this :
which is also correct. Although, it would have looked better with BRACKETS.


* Do not neglect the BRACKETS, as indicated in this question partly because it does not confuse the person and also because it looks better =).

* The negative part of the expression can be anywhere. It doesn't have to be at the end of the logarithm. Mr. K. pointed out that it is clever to solve and simplify in only a few steps but also it will help a novice to understand what's happening through the display of the intermediate step.

TAKE NOTE: Proportinal reasoning does not work with EXPONENTS because exponential functions increase exponentially as oppose to straight lines.

Next, we were given values of some logarithms with specific values for the following logarithms.
GIVEN : loga 2 = 0.3562; loga 3 = 0.5646; loga 5 = 0.8271


  • Expand logarithms using the Law of Logarithm.
  • Re-write the expanded logarithms using the Law of Logarithm.
  • We now see either loga2, loga3, or loga5. We already know those values so we just substitute.
  • Then perform the required operation.



  • First line, let unknown value be x.
  • Definition of a logarithm - logbasePOWER = Exponent.
  • Indication of log10 if there is no base. It doesn't show because "10 is conveniently programmed in our calculators" - Mr. K.
  • Isolate x on one side (either left or right; there is no left-hand prejudice rule).
  • You end up with an equation in the CHANGE OF BASE FORMULA.
  • x is approiximately 1.5850.


Change of Base Formula : ( c is the new base of 10 )
logba = logca / logcb


--- So then Mr K. began, "WHEN WE SOLVE LOGARITHMS, WE USE 4 DECIMAL APPROXIMATION." Then he reiterated that concept so many times in different ways because he knew a few of us will always forget it. He reminded us that if did not round off up to 4 decimals, it would be a full mark deduction because it is a concept error. Then, we tackled questions involving approximate values of logarithms. Although, it was unfortunate that the class had to end.

AFTERNOON CLASS

--- Fairly frustrated with how the experiment went in Chemistry, we went to Pre-Cal. From there, we carried on with the problems given to us in period 1.





  • Rewrite the power using the LAW OF LOGARITHM with a base of 10
  • Bring down the exponent using
  • Isolate x and using the CHANGE OF BASE FORMULA
  • Make sure subtracting 1 is done after the division.

--- Up to this point, everyone was dedicated in solving the problems on the SMARTBOARD. It was that time when Mr. K. showed us another unbelievable project. The project contained a lot of animations and was incredibly done. Again, we were amazed and at the same time, pressured due to the fact that our projects must meet that level. Well, not really. But still, who didn't feel pressured at that time? Yeah Yeah, maybe a few people didn't but I was! Then, we solved this problem.


--- Mr. K. also quoted, "There's more than one way to skin a cat, but don't literally skin a cat because that's mean." He meant that there's another way that we could have solved logarithmic problems. In the previous question, we could have multiplied both sides by 1/2, resulting in the isolation of 3x on the left hand side and 5 / 2 on the right hand side. From there, we can easily use the CHANGE OF BASE FORMULA, and skipped two intermediate steps. Also don't forget to apply brackets around the numerator.



  • It is an intermediate step.
  • Bring down the exponents using logbMk = klogbM.
  • Subtract two from both sides.
  • Use the distribution process to expand (x+1) log6.
  • Isolate all the terms with x.
  • Factor x out.
  • Multiply both side by 1 / (log3 - log6) then evaluate.
  • Shortcuts or other ways to do the problem.

  • Bring down the exponents using logbMk = klogbM.
  • Isolate like terms.
  • Factor x out.
  • Evaluate.

  • Use logb MN = logb M + logb N to combine all xs.
  • Use FOIL, then rewrite the equation in logbasePOWER = Exponent form.
  • Evaluate 53 then subtract 125 from both sides.
  • Since the quadratic equation does not factor nicely, we will use the quadratic formula.
  • Then, x = 6.8226 and x = -6.1560 .
  • However we cannot use the negative value because it is an extraneous solution. Although, we have to include both solutions and we have to STATE THAT THE NEGATIVE VALUE IS REJECTED.

The reason for the rejection of the negative solution is that:

--- LogbaseARGUMENT = Exponent or logbasePOWER = Exponent, if and only if BASEexponent = POWER. The base has to be positive in that law of logarithm. If the base is negative, a funky dotted graph is produced. If we accept the negative number, then logbase negative number = Exponent. Therefore, this statement must also be true: BASEexponent = negative number. However, it's not correct because if we substitute -6.1560, we will have:

log5[3(-6.1560+1)(-6.1560-1)] = 3

log5(125) = 3

53 = 125 (but, 125 isn't negative at all like what we had earlier : If logbase negative number = Exponent, then BASEexponent = negative number. Obviously, we are not given a negative number as a power at all.

--- On the other hand, if we substitute a positive number, it will all work out!

--- This concludes the day's lessons. It was so intense having to cover two-days worth of material. Now, We have to do Exercises 22 - 24 by Monday. Just a reminder to all of you that he will not be in class on Monday and Tuesday. So then, after the class, a few people styeyd behind to find out what our marks so are. And continuing on with the second cycle of the Scribe Routine, the next scribe will be ROBERT!

10 comments:

e said...

Nice job, Vincent. I was about to ask you a question, when I realized that you had asked it yourself:
However we cannot use the negative value because it is an extraneous solution. Although, we have to include both solutions and we have to STATE THAT THE NEGATIVE VALUE IS REJECTED. I'm not really sure of what the real reason is. Maybe you guys can tell me.
I am looking forward to the answer. Maybe looking at the first line of the problem would be helpful. Thoughts?

Lani said...

Hi Vincent,

What a great scribe! Your title is creative and the opening graphic really pulls your reader in. That you have color coded your annotations with the arrows on the slides adds to the understanding of the processes.

Congratulations on a job well done!

Best,
Lani

aichelle s. said...

good job Vincent I like your constant reminders...your arrows and circles and use of colour it's an excellent scribe!

MilesM said...

Wow... Awesome Post vincent... Te color makes it easy to understand... Great use of the marquee thing... it looked so cool/// :P

VincentR said...

Hi e! I hope that satisfies your question. I got a little help from MR. K. and I gotta say, when I fully concentrated on what he hinted at, I actually got it! 125 isn't a negative number if the binomial was evaluated, and it totally contradicts the statement LOGbase NEGATIVE NUMBER = POWER -> base^exponent = negative number. Thanks for making me think about it. I learned a lot just because of that!

e said...

Thank you for thinking about my questions. And I think you are on the right track. I am little confused about the last couple of lines. You have all the right information, but it seems you rushed a little right here:
However, it's not correct because if we substitute -6.1560, we will have:

log5[3(-6.1560+1)(-6.1560-1)] = 3


If you put in -6.1560 right there, there actually is no problem, as when you evaluate the product you get a positive number, just like you did in the next line. But try putting that number in for x in the first line of the initial problem. Then you will encounter negative numbers where you shouldn't. Try it.

Logarithms are very important functions, and often give students some headache. What you and Mr. K were secretly discussing was the domain of the logarithm function. Can you tell us what the domain and range of logaritm is? Think about what logarithm is to exponential function and try using that. Do you see how that relates to the problem that we were discussing?

VincentR said...

Hi e! Thanks for the suggestion. As for your other question, this is what I have so far:
'Domain and range for logarithms vary because they both rely on the transformations applied to the equation. Also, logarithms have asymptotes therefore, range and domain must be affected by the asymptote. Also, sticking in a negative sign in front of a logarithm will reflect it over the y-axis.' I'm not entirely sure of this concept. Please keep on squeezing it out of me. Maybe I'm not that far away from nailing the answer.

Mr. Kuropatwa said...

Vincent, you're doing some good thinking here but you're thinking about the function in terms of all the transformations that "could" be applied to it. What if no transformations are applied ...

What is the domain and range of the logarithm function?

Is the asymptote of the logarithm function connected to the domain, range or both? You said both. Are you sure? Why is this?

e has suggested several times that you look at the original equation and substitute the negative result into that. You've tried substituting that value in everywhere except there. Take e's advice, explicitly, and see what happens. This also has something to do with the question about domain and range above.

Like I said, you're doing good thinking here, don't stop now. ;-)

«Craig» said...

Hall of Fame!!! It took the guy five hours!!! Just give it to him!

Seriously, it is so detailed. Every part of the class and the things that we learned are so carefully explained. Visuals are more than visuals. They are tools for learning and it is great how he elaborated on the slides. :)

e said...

Vincent,

I have an image of me wringing a little rag trying to squeeze some math out of it :) You're funny. I agree with Mr. K: keep on thinking. You should involve some of your classmates, too. Here is another suggestion that may help. Can you graph for me the following function
v(x)= log(x),
where the base is 10 (and you got it, v is for vincent). That may help with domain and range.