SsSsSs-Spring-cribe

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:

log_basePOWER = Exponent

log_b MN = log_b M + log_b N

log_b (M/N) = log_b M - log_b 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 : log_a 2 = 0.3562; log_a 3 = 0.5646; log_a 5 = 0.8271

• Expand logarithms using the Law of Logarithm.
• Re-write the expanded logarithms using the Law of Logarithm.
• We now see either log_a2, log_a3, or log_a5. 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 - log_basePOWER = 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 )
log_ba = log_ca / log_cb

--- 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 3^x 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 log_bM^k = klog_bM.
• 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 log_b MN = log_b M + log_b N to combine all xs.
• Use FOIL, then rewrite the equation in log_basePOWER = 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:

--- Log_baseARGUMENT = Exponent or log_basePOWER = Exponent, if and only if BASE^exponent = 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 log_base negative number = Exponent. Therefore, this statement must also be true: BASE^exponent = 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 log_base negative number = Exponent, then BASE^exponent = 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!