Question: Simplify: (log210)(log48-log3)
Solution: (log10/log2)(log16)
=(1/log2)(log2^4)
=4log2/log2
=4
Wednesday, November 25, 2009
Test Question
1st question (2 marks)
Convert expression into a single expression logarithm:
½ln(x) - 2 ln(x-1) - 1/3 ln (x²+1)
=ln √(x) - ln(x-1)² - ln ³√(x²+1)
= ln (√x)/ [ (x-1)² * ³√(x²+1) ] <--- Log Law #2.
surprisingly, thats all you have to do for that question (simplify)
2nd question (1 mark)
Express log 6 1/216= -3 in exponential form
6^-3= 1/216
3rd question (1 mark)
Evaluate ln ( 1/ 4√e)
*multiple choice here guys..
a) ¼
b) -¼
c) -4
d) 4
Convert expression into a single expression logarithm:
½ln(x) - 2 ln(x-1) - 1/3 ln (x²+1)
=ln √(x) - ln(x-1)² - ln ³√(x²+1)
= ln (√x)/ [ (x-1)² * ³√(x²+1) ] <--- Log Law #2.
surprisingly, thats all you have to do for that question (simplify)
2nd question (1 mark)
Express log 6 1/216= -3 in exponential form
6^-3= 1/216
3rd question (1 mark)
Evaluate ln ( 1/ 4√e)
*multiple choice here guys..
a) ¼
b) -¼
c) -4
d) 4
Thursday Test Question
This is my test question:
Solve for x: ln6+ln(x−2)=−2
(Express your answer correct to 3 decimal places.)
This question is worth 3 marks as well
Here is the method I used to get the answer:
ln[6(x-2)]=-2
ln(6x-12)=-2 (1 mark here)
(e^-2)=6x-12
6x=(e^-2)+12
x=(e^-2)+12/6 (1 mark here)
x=2.022 556
x=2.023 (1 mark here)
(that's where the marking guide said the marks should be awarded)
I would suggest to everyone that you go over all the test questions mentioned on here. This way you know what you will have to answer and how to do it. Good Luck to everyone!
Solve for x: ln6+ln(x−2)=−2
(Express your answer correct to 3 decimal places.)
This question is worth 3 marks as well
Here is the method I used to get the answer:
ln[6(x-2)]=-2
ln(6x-12)=-2 (1 mark here)
(e^-2)=6x-12
6x=(e^-2)+12
x=(e^-2)+12/6 (1 mark here)
x=2.022 556
x=2.023 (1 mark here)
(that's where the marking guide said the marks should be awarded)
I would suggest to everyone that you go over all the test questions mentioned on here. This way you know what you will have to answer and how to do it. Good Luck to everyone!
Transformations Questions
Questions: #13 - 2 marks (1 per letter)
#14 - 3 marks (1 per letter)
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20QUESTIONS.pdf
Answers Link:
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20ANSWERS.pdf
#14 - 3 marks (1 per letter)
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20QUESTIONS.pdf
Answers Link:
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20ANSWERS.pdf
Test Questions
Here are my questions for our test tomorrow:
1) Point P is a point on the terminal arm of an angle Θ in standard position. If cscΘ<0 and sin theta=-15/17, deterine the value of the other five trigometric functions.
2)P(Θ) is a point on the unit circle. If secΘ<0 and tan(theta)<0, and the coordinates for P are (x, 2root6/7)
a. Find the value of x
b.State the coordinates of P(Θ+pi)
Solutions:
1)step 1: Determine quadrat-(quadrat III)
step2: Find sinΘ=1-(-8/17)squared=-15/17
Step3: Determine other 4 trig. values
sec=-17/8
csc=-17/15
tan=15/8
cot=8/15
2) a.
step1: determine quadrat-(quadrat 2)
step 2:1-(2√6/7)²=-5/7
b.step1: determine quadrat-(quadrat 4)
step 2: determine integer values for x and y: (5/7, -2root6)
1) Point P is a point on the terminal arm of an angle Θ in standard position. If cscΘ<0 and sin theta=-15/17, deterine the value of the other five trigometric functions.
2)P(Θ) is a point on the unit circle. If secΘ<0 and tan(theta)<0, and the coordinates for P are (x, 2root6/7)
a. Find the value of x
b.State the coordinates of P(Θ+pi)
Solutions:
1)step 1: Determine quadrat-(quadrat III)
step2: Find sinΘ=1-(-8/17)squared=-15/17
Step3: Determine other 4 trig. values
sec=-17/8
csc=-17/15
tan=15/8
cot=8/15
2) a.
step1: determine quadrat-(quadrat 2)
step 2:1-(2√6/7)²=-5/7
b.step1: determine quadrat-(quadrat 4)
step 2: determine integer values for x and y: (5/7, -2root6)
My question.
These questions are pulled from the Mickelson book.The solutions can be found on pages 198 and 199.
Page: 179
Number:16)
Letter: c)
Evaluate(worth one mark):
10! 4! / 8! 6!
= (1*2*3*4*5*6*7*8*9*10)(1*2*3*4) /
(1*2*3*4*5*6*7*8) (1*2*3*4*5*6)
= 9 * 10 /
5 * 6
90 / 30
= 3
Second question (page 180, number 23) (worth two marks):
How many different groups of four letters can be made from the letters A, B, C, D, E, and F if the letters can only be used once?
P(6, 4) = 6! / (6 - 4)! = 6! / 2! =
1 * 2 * 3 * 4 * 5 * 6 /
1 * 2
= 3 * 4* 5 * 6
= 360
We can create 360 groups of four letters from A, B, C, D, E, and F if the letters can only be used once.
Tuesday, November 24, 2009
TEST QUESTION!
HERE WE GO GUYS...QUESTION ON THE FUNDAMENTAL COUNTING PRINCIPLE (FCP)
This is out of the Mickelson (hmm i always thought it was nickleson) book. pg 176.
There are 4 roads between cities A and B, and 3 roads between cities B and C.
a) How many ways can a person travel from A to C through city B??
(4)(3) = 12
b) How many ways can a person make a round trip from A to C and back to A, through city B?
(4)(3)(3)(4) = 144
c) How many ways can a person make a round trip from A to C and back to A through city B, without using any road twice?
(4)(3)(2)(3) = 72
Alright, hope that helped and you all do great
This is out of the Mickelson (hmm i always thought it was nickleson) book. pg 176.
There are 4 roads between cities A and B, and 3 roads between cities B and C.
a) How many ways can a person travel from A to C through city B??
(4)(3) = 12
b) How many ways can a person make a round trip from A to C and back to A, through city B?
(4)(3)(3)(4) = 144
c) How many ways can a person make a round trip from A to C and back to A through city B, without using any road twice?
(4)(3)(2)(3) = 72
Alright, hope that helped and you all do great
Brett's Test Question
Question:
1. a) How many 6 letter arrangements can be made using the letters M A T T E R ? Give your answer as a whole number. (1 mark)
b) How many 6 letter arrangements end with T T? Give your answer as a whole number. (1 mark)
c) How many 6 letter arrangements can be made if the letters M T T R must be together? Give your answer as a whole number. (2 marks)
Solutions:
1. a) 6!/2!= 360
b) _ _ _ _ T T
4!= 24
c) (3!) (4!/2!) = 72
**3 groups, 4 consonants, 2 t's
GOOD LUCK THURSDAY!
1. a) How many 6 letter arrangements can be made using the letters M A T T E R ? Give your answer as a whole number. (1 mark)
b) How many 6 letter arrangements end with T T? Give your answer as a whole number. (1 mark)
c) How many 6 letter arrangements can be made if the letters M T T R must be together? Give your answer as a whole number. (2 marks)
Solutions:
1. a) 6!/2!= 360
b) _ _ _ _ T T
4!= 24
c) (3!) (4!/2!) = 72
**3 groups, 4 consonants, 2 t's
GOOD LUCK THURSDAY!
My test question!
I just got this from accelerated math, so i'll write it out for you as opposed to posting a link!
(Worth 2 marks)
Write as a logarithm of a single expression:
2log5x + 3log5(x+6)
Answer:
= log5x^2 + log5(x+6)^3
= log5x^2(x+6)^3
Taaa Daaaa!
(Worth 2 marks)
Write as a logarithm of a single expression:
2log5x + 3log5(x+6)
Answer:
= log5x^2 + log5(x+6)^3
= log5x^2(x+6)^3
Taaa Daaaa!
Test Question for Nov 26 2009
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20QUESTIONS.pdf
Question #2
a) Worth 2 marks
b) Worth 2 marks
c) Worth 2 marks
Question #11 Part A, B, C and D
Answers can be found here...
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20ANSWERS.pdf
Question #2
a) Worth 2 marks
b) Worth 2 marks
c) Worth 2 marks
Question #11 Part A, B, C and D
Answers can be found here...
http://www.bmlc.ca/Math40S/Pre-Calculus%20Math%2040s%20Standards%20Test%20-%20Transformations%20ANSWERS.pdf
my test questions
HEYYYYYYYY YOUUUUUUU GUYYYYYYYY'S
If you would like the advanced edge on tomorrows upcoming test, here it is.
I Picked two questions, one from exercise 3 and one from exercise 5.
from exercise 3, I picked question 1, part e), which was
find the exact value of tan(7pi/4), show work, ie. unit circle
from exercise 5, question 1
Solve for thaeta where the domain is the set of all reals.
sin thaeta=1/2
answer is pi/6+2kpi where k is an integer
and
5 pi/6 + 2kpi where k is an integer
dont forget to show how you got those answers. ie unit circle
If you would like the advanced edge on tomorrows upcoming test, here it is.
I Picked two questions, one from exercise 3 and one from exercise 5.
from exercise 3, I picked question 1, part e), which was
find the exact value of tan(7pi/4), show work, ie. unit circle
from exercise 5, question 1
Solve for thaeta where the domain is the set of all reals.
sin thaeta=1/2
answer is pi/6+2kpi where k is an integer
and
5 pi/6 + 2kpi where k is an integer
dont forget to show how you got those answers. ie unit circle
In class today Mr. Max gave us more examples on Perms and coms. The first example that we were given showed us how to create possible cases for one question, but after creating the possible cases you combine them together to get the final answer.
Next, we were given the term and definition: Sample Space- some display, rendition, showing all probability outcomes. "has everything" Scrabble was used as the example to explain sample space, which is shown below.
Test has now been moved to Thursday everyone, and there is tutoring today after school with Mr.Max. The Test questions should be posted on to the blog by wither today or tomorrow, where we will be able to see the problems and solutions. Good luck everyone!
Next, we were given the term and definition: Sample Space- some display, rendition, showing all probability outcomes. "has everything" Scrabble was used as the example to explain sample space, which is shown below.
Test has now been moved to Thursday everyone, and there is tutoring today after school with Mr.Max. The Test questions should be posted on to the blog by wither today or tomorrow, where we will be able to see the problems and solutions. Good luck everyone!
Tuesday November 24, 2009- More Perms and Coms!
First of all, TEST MOVEMENT! Instead of Wednesday, we're moving this to Thursday.
What happens if the orderings of perms/coms are circular? i.e) people around a table; bracelets; necklaces; basically things that require objects in sequence on a round form of something.
ex) How many ways can you sit 4 people around a table?
*moving to the cafeteria to do some shananigans around the tables.
-How many things are actually moving?
ANSWER- 3 things are actually moving, which makes the equation 3!
Final answer- 3!= 6 ways
FORMULA: (n-1)!, where n= number of things to order in a circular way.
ex) Regarding this necklace, how many ways can we make this thing unique?
Thinking the same as the previous example? If so, that is incorrect. The 3D object can be flipped around, as shown in the picture. FORMULA: (n-1)!/2
ex) 5 people around a circular table
a) if person A and person B have to sit together?
{(n-1)!} *2 <--- because they can switch around (A-B, B-A) b) if person A and person B can not sit together? FORMULA: (All ways)-(undesired ways)= Answer.
4! - 3!2! = 12 ways
c) person A and person B together; person C and person D together.
2! (from the n-1 ! formula); 2! ways to organize A/B; 2! ways to organize C/D
2! * 2! *2!= 8 ways.
Ordering identical things-
ex) "How many different 'words' are possible with these letters?"
Z O O M
*thinking the answer is 4! ? 24 ways? Wrong, again!
MOZO* is the same as MO*ZO..
FORMULA: total number of ways / "total number of identicals"! = desired ways
ex) S T A T I S T I C S
10 !/ {(3!)(3!)(2!) = 50400 UNIQUE words
What happens if the orderings of perms/coms are circular? i.e) people around a table; bracelets; necklaces; basically things that require objects in sequence on a round form of something.
ex) How many ways can you sit 4 people around a table?
*moving to the cafeteria to do some shananigans around the tables.
-How many things are actually moving?
ANSWER- 3 things are actually moving, which makes the equation 3!
Final answer- 3!= 6 ways
FORMULA: (n-1)!, where n= number of things to order in a circular way.
ex) Regarding this necklace, how many ways can we make this thing unique?
Thinking the same as the previous example? If so, that is incorrect. The 3D object can be flipped around, as shown in the picture. FORMULA: (n-1)!/2
ex) 5 people around a circular table
a) if person A and person B have to sit together?
{(n-1)!} *2 <--- because they can switch around (A-B, B-A) b) if person A and person B can not sit together? FORMULA: (All ways)-(undesired ways)= Answer.
4! - 3!2! = 12 ways
c) person A and person B together; person C and person D together.
2! (from the n-1 ! formula); 2! ways to organize A/B; 2! ways to organize C/D
2! * 2! *2!= 8 ways.
Ordering identical things-
ex) "How many different 'words' are possible with these letters?"
Z O O M
*thinking the answer is 4! ? 24 ways? Wrong, again!
MOZO* is the same as MO*ZO..
FORMULA: total number of ways / "total number of identicals"! = desired ways
ex) S T A T I S T I C S
10 !/ {(3!)(3!)(2!) = 50400 UNIQUE words
Sunday, November 22, 2009
Friday Nov 20. Work Period.
Friday we had a work period to catch up on Accelerated Math, Cumulative Exercises and to work on our 2 questions of our choice picked from Old exams, Cummulitave Excercises, Accel Math, ect to be submit for the test on Wednesday.
It would be wise to be into the Probility on Acell Math and have the following Cummulative Exercises completed prior to Wednesdays test...
**Up to and including**
Ex 1 #1-12
Ex 2 # 1-12
Ex 3 # 1-7
Ex 4 #1-8
Ex 5 #1-12
Ex 6 #1-16
Ex 7 # 1-10
Ex 8 #1-10
Ex 9 # 1-14
Ex 10 # 1-6
Ex 11 # 1-12
Ex 12 # 1-10
Ex 13 #1-10
Ex 14 #1-11
Ex 15 # 1-14
Ex 16 # 1, 2, 4, 6-13
Ex 17 # 1-12
Ex 18 # 1-18
Ex 19 # 1-10
Ex 20 # 1-13
Ex 21 # 1-12
Ex 22 #1-10
Ex 23 # 1-12
Ex 24 # 1-10
Ex 25 # 1-15
Ex 26 # 1-16
Ex 27 # 1-10
Ex 28 # 1-10
Ex 29 # 1-10
It would be wise to be into the Probility on Acell Math and have the following Cummulative Exercises completed prior to Wednesdays test...
**Up to and including**
Ex 1 #1-12
Ex 2 # 1-12
Ex 3 # 1-7
Ex 4 #1-8
Ex 5 #1-12
Ex 6 #1-16
Ex 7 # 1-10
Ex 8 #1-10
Ex 9 # 1-14
Ex 10 # 1-6
Ex 11 # 1-12
Ex 12 # 1-10
Ex 13 #1-10
Ex 14 #1-11
Ex 15 # 1-14
Ex 16 # 1, 2, 4, 6-13
Ex 17 # 1-12
Ex 18 # 1-18
Ex 19 # 1-10
Ex 20 # 1-13
Ex 21 # 1-12
Ex 22 #1-10
Ex 23 # 1-12
Ex 24 # 1-10
Ex 25 # 1-15
Ex 26 # 1-16
Ex 27 # 1-10
Ex 28 # 1-10
Ex 29 # 1-10
Friday, November 20, 2009
Deeper Thinking...
Yesterday in class Mr. Max had us do a different style of assignment. We were to read an article online (link is below in his post) and write our own blog post based on our thoughts on the topic. It was about an agreement made between 2 students and a teacher about having a No Homework agreement where the 2 students would get, well, no homework. Ever.
So, your homework is to write a blog post on what you think about having no homework. Ironic?
We spent the rest of the block working on accelerated math. Try to get that done guys, I know I myself am super far behind and its just taking me ages to get caught up, but once I complete an objective I really do feel like I know what I'm doing for the most part.
REMINDER: Test is next Wednesday, so lets get our study on guys!
So, your homework is to write a blog post on what you think about having no homework. Ironic?
We spent the rest of the block working on accelerated math. Try to get that done guys, I know I myself am super far behind and its just taking me ages to get caught up, but once I complete an objective I really do feel like I know what I'm doing for the most part.
REMINDER: Test is next Wednesday, so lets get our study on guys!
Pilot Exam
I forgot to post this when I scribed for the class on Monday. For anyone who missed Monday's class, we were told that there is going to be a pilot exam (practice exam) for our Grade 12 Pre-Calculus exam after holiday break. This takes place at 12:30 on January 14th. I think this is a great opportunity to get a feel for what the real exam is like as well as to find find out where we need to improve before the actual exam. It is going to take place in Mr. Maksymchuk's room.
Thursday, November 19, 2009
Perms and Coms
Yesterdays class, we learned more about perms and coms. Permulations are the number of possible orderings of a set, provided that the order of set elements matters. An example would be 281-0563 218-0563 560-3812 or books on a shelf, locker combos. The BEST EXAMPLE EVER of a perm, is create 5 letter words from the alphabet no repeats. So you pick any 5 letters, and then you would times 26(because there is 26 letters in the alphabet) then 25 because you used a letter, then 24 b/c you used another letter...etc. you get 7893 600. At that point we learned the formula : 26!/21!=n!/(n-r)! n being the number of items to select from and r being selecting items r at 2 time. "n things, permulated 5 at a time". The second part of THE BEST EXAMPLE EVER is: pizzas. you have peperoni pinapple and ham, either order you put it in, its still the same pizza.
To Homework or Not? That is the question?
http://www.cbc.ca/canada/calgary/story/2009/11/18/calgary-homework-school-students.html
So what do you think? As young adults enrolled in the final year of public education, specifically in mathematics, which side of this 'argument' do you line up on?
Discussion prompts?
Would you like a no-homework agreement?
Would you benefit from a no-homework agreement?
Would you learn enough with no homework?
Do you have enough time in school to both learn and practice?
Is the subject area a consideration when discussing such agreements?
I really don't know the answers, but inspirationally, I know you do. I look forward to reading your responses. Thank you in advance for informing both public education practice generally and specific decisions made in room 204 at the SVRSS.
So what do you think? As young adults enrolled in the final year of public education, specifically in mathematics, which side of this 'argument' do you line up on?
Discussion prompts?
Would you like a no-homework agreement?
Would you benefit from a no-homework agreement?
Would you learn enough with no homework?
Do you have enough time in school to both learn and practice?
Is the subject area a consideration when discussing such agreements?
I really don't know the answers, but inspirationally, I know you do. I look forward to reading your responses. Thank you in advance for informing both public education practice generally and specific decisions made in room 204 at the SVRSS.
Monday, November 16, 2009
Logarithmic Word Problems
Today in class we did logarithmic word problems. For the most part, it is just review from Grade 11 Pre-Calc. The first two images are a little small because of the formatting.
The first thing we did dealt with was radioactive decay and half lives. The first question had to do with finding the amount of substance left after a determined amount of time; In our case, it was 3 years. So, all you have to do is substitute your X value for 3 and from there it becomes calculator work.
The next section had to do with population increase. However, this formula can be used for anything that has to do with exponential growth. In this case, it ts used to show growth rates of bacteria. Also for this one, you just substitute in your values and solve.
Again, we are asked to find the half life of a substance. This question is just like the last one other than it is a different formula. You use natural logarithms to get rid of your e value, use the logarithmic laws, get x by itself, and solve with your calculator.
The first thing we did dealt with was radioactive decay and half lives. The first question had to do with finding the amount of substance left after a determined amount of time; In our case, it was 3 years. So, all you have to do is substitute your X value for 3 and from there it becomes calculator work.
The next question involves a little more work and use of the logarithmic laws. We were asked to find the half life of the substance. What is asked here is what amount of time does it take for the substance to decay to half its original mass. So, we're solving for X. We know that the original mass was 80g so the mass at its half life would be 40g. Our Y value is now 40. Next, you multiply everything by ln. This is usually beneficial because out original question has the variable e in it so there is a good chance that we can get rid of it by using ln. Remember, the ln(80e^-0.2x) is actually ln80+ln(e)^-0.2x since it is a law. Next, we bring down the x value (another law) so we can factor it and get x by itself. From here, it becomes calculator work.
Out next section had to do with compound interest. Our first question had to do with finding the value of the investment after 1 year. All you have to do is substitute 1 in for the t value. From there it becomes calculator work.
The next question had to do with finding the value of the investment after 10 years. Again, just substitute 10 in for the t value and take it from there.
The last question had to do with finding the interest after 10 years. All you have to do for this is find the value of the investment after 10 years and subtract it from the original investment ($5000)
This formula shows what the interest would be if it was compounded all the time as opposed to just monthly or annually. How this formula was derived is that when the interest reaches a very high number, it gets very close to the value e. So the formula (1+r/n)^n simply becomes e^r. The reason why banks don't use this way of compounding interest is because it means more money for the consumer. The use of this formula is strait forward; You just substitute in your values and solve with your calculator.
The next section had to do with population increase. However, this formula can be used for anything that has to do with exponential growth. In this case, it ts used to show growth rates of bacteria. Also for this one, you just substitute in your values and solve.
Again, we are asked to find the half life of a substance. This question is just like the last one other than it is a different formula. You use natural logarithms to get rid of your e value, use the logarithmic laws, get x by itself, and solve with your calculator.
We were assigned 1-16 on exercise 26 and 1-10 on exercise 27.
I think we're assigned up to objective 31 on our accelerated math to be done before our test next Wednesday.
Friday, November 13, 2009
Thursday, November 12, 2009
On Thursday we had a substitute teacher, so we didn't learn anything new. We worked on assigned work such as accelerated math and our student exercises. I was really having trouble with this unit until I found a website that explains the concept of logarithms and it gives you several example questions. Also if you are having trouble with things from the last units you can find all the previous topics on the website and they are explained fully. The website is http://www.brightstorm.com/. When you get to the site you just click on the math link then it will give you a list of topics, select pre-calculus, then a list of all the units we took or will be taking this year pops up. Then select a video.
"Success is a peace of mind which is a direct result of self-satisfaction in knowing you did your best to become the best you are capable of becoming."- John Wooden
"Success is a peace of mind which is a direct result of self-satisfaction in knowing you did your best to become the best you are capable of becoming."- John Wooden
natural logarithms(-notations-equations-and applications)
REMEMBER! loge = ln
OK, for this first slide he used 3 logarithmic identities to form the single logarithm
1. ln(x-1) + 3ln(x+3) becoming the numerator(x-1)(x+3)^3 from
the identity log e (MN)= log e M + log e N
2.from (ln(x-1)+3ln(x+3))-1/2ln(x^2+2) he got the denominator Square Root of (x^2+2) which is the same as(x^2+2)^1/2 from the identity log e (M/N) = log e M - log e N
3. and the third identity is log e (M^N) = N log e M which you can see done with both the numerator and denominator
the only thing i think i need to explain with this one is that if both sides of an equation have loge
in front we know we can take it away, correct? so i guess the opposite can be done as well where we add in loge, which is seen done in the second line (the red ink).
(remember! loge = ln)!
*lne=1
this slide uses the third identity mentioned in the first slide, and takes away ln from both sides(as mentioned in slide two), other than that i think it is self explanatory.
in this slide he adds ln to both slides making lne on the left side which previously stated equals one(see lime green * in above slides explination).
and again he used the third identity mentioned in slide 1.
Hope this helped you in anyway, shape, or for possible
Mr. Maksymchuk gave use another one of his shirt ideas this one was based on the Pink Floyd album cover the dark side of the moon with the rainbow and prisms, but his shirt would say Pink Freud. I didn't get the reference but maybe you did.
I later googled it and it was referanced to Sigmund Freud who was a phsycoanalysis who was best know for his theories of the unconcious mind and the defence mechanism of repression.
after all that i still didn't completely get it, oh well, my loss I guess
Tuesday, November 10, 2009
Approximating "e" and The Natural Logarithms (ln)
Homework: Exercise 25 #1-15
Also, remember to vote for which day you think our next test should be!
We were also taught about Natural Exponential Functions and what they look like.
The graph of f(x) = e^x has these properties: (Important to know and understand)
We also sketched some Exponential Graphs on Graphmatica
a)y= e^x-2
Compared to y=e^x, this graph is shifted 2 units down.
b)y=-e^-x
Compared to y=e^x, this graph is flipped over the y-axis, then the x- axis.
c)y= abs(e^x-1)
Compared to y=e^x, this graph is moved 1 unit down, then taking the absolute value of it causes the negative part of the graph to be bounced up and over onto the positive side of the x-axis.
If you compare graphs a, b, and c to the original equation (y=e^x) on Graphmatica, being able to see the actual changes really help. Also, if you know the approximate look of the original equation, you can just apply what we learned in our transformation unit and visualize where to move the graph by how much and so on. (It's pretty neat how all math builds on each other for the most part!)
Also, remember to vote for which day you think our next test should be!
Mr. Maks told us that the general idea of the lesson was that: loge=ln, in the case that e~2.71828...
We used the formulas: f(n) = 1+1/n , g(n)= (1+1/n)^n (for values n E I)
and Mr. Maks showed us, using Excel, the comparison between the two formulas using numbers between 1 and 1000.
and Mr. Maks showed us, using Excel, the comparison between the two formulas using numbers between 1 and 1000.
We were also taught about Natural Exponential Functions and what they look like.
The graph of f(x) = e^x has these properties: (Important to know and understand)
- D= R (reals)
- R= (y>0)
- increases on its entire interval (moves upwards to the right)
- concave up
- one to one (** means it has an inverse**) ; if e^x1 = e^x2 (therefore, x1 = x2)
- 0<> 1 for x> 0
This relates back to the important equations Mr. Maks gave us yesterday of:
e^1nx=x and f(f^-1(x))=x (try them out on your calculator, they actually work!)
e^1nx=x and f(f^-1(x))=x (try them out on your calculator, they actually work!)
We also sketched some Exponential Graphs on Graphmatica
a)y= e^x-2
Compared to y=e^x, this graph is shifted 2 units down.
b)y=-e^-x
Compared to y=e^x, this graph is flipped over the y-axis, then the x- axis.
c)y= abs(e^x-1)
Compared to y=e^x, this graph is moved 1 unit down, then taking the absolute value of it causes the negative part of the graph to be bounced up and over onto the positive side of the x-axis.
If you compare graphs a, b, and c to the original equation (y=e^x) on Graphmatica, being able to see the actual changes really help. Also, if you know the approximate look of the original equation, you can just apply what we learned in our transformation unit and visualize where to move the graph by how much and so on. (It's pretty neat how all math builds on each other for the most part!)
Also, like Mr. Maks told us there's really great explinations and examples of logs in the Mickelson book starting on page 37.
Monday, November 9, 2009
Monday, November 9th: change of base and solving exponential equations.
Today we learned about the change of base formula; also, it was semester day 40, so we are halfway in. Mr. Max believes that we should work hard at our accelerated math and do the relevant sheets before the test. If we were to do that, we'd do much better on the test. If you're a little behind on the sheets for our upcoming test, that's fine. You could always get caught up before the next test, and keeping on top of accelerated math would make things less stressful.
logbn = logan / logab
As long as we follow the pattern of the formula, you can make it workable on any base you choose.
given y = log2 3
can be rewritten as:
y = log10 3 / log10 2
How to we evaluate? We can either use our calculator and get y~ 1.585 . . .
Or, we can use logs. (No kidding, eh!)
y = log2 3
change to an exponential expression.
2^y = 3
then we apply the def of logs: log(2^y) = log3
y = log10(3) / log10(2)
Solving exponential equation.
Solve:
2(3^x) = 5
CAVEAT! (A latin word that means "warning" or "beware"):
log (x + 3) != (not equal to) logx + log3
instead, it becomes:
log(2(3^x)) = log5
log2 + log3^x = log5
log2 + x * log3 = log5
x * log3 = log5 - log2
x = log5 - log2 / log3 (at this point, we cannot continue without a calculator)
x = .8340437671
to check out answer: 2 * 3 ^ .8340437671 = 5
We just solved our very first exponential equation!
Let's solve another one:
19^(x - 5) = 3^(x+2)
(Note: Our reason for doing the log is to bring the exponents down.)
log 19 ^ (x - 5) = log 3^(x+2)
x * log19 - 5 * log19 = x * log3 + 2 * log 3
(x - 5) * log 19 = (x + 2) * log 3
x * log19 - 5 * log19 = x * log3 + 2 * log3
x * (log19 - log3) = 2 * log3 + 5 * log19
x = 2 * log3 + 5 * log19 / (log19 - log3) (now we use our calculator)
x ~ 9.166
log2 (x - 2) + log2(x) = log23
log2(x - 2)(x) = log23
x - 2(x) = 3 (we can remove the logs since they have the same base)
x ^ 2 - 2x - 3
(x + 1)(x - 3) = 0
x = -1, x = 3
STOP!
x = -1 doesn't solve the equation.
x = 3 works, though
log5(3x + 1) + log5(x - 3) = 3
log5[(3x+1)(x-3)] = 3
(use definition of logs:
5 ^ 3 = (3x + 1)(x - 3)
logab = n
a^n = b)
125 = 3x^2 - 8x -3
0 = 3x^2 - 8x - 128
(384: 16 and 24 as factors)
0 = 3x^2 - 24x + 16x - 128
0 = 3x(x-8) + 16(x-8)
0 = (3x + 16) (x - 8)
3x + 16 = 0, x - 8 = 0
x = -16/3 and x = 8
check for error:
- 16 /3 isn't a solution because after we plug it in, there is no value that 5 ^x that would be equal to a negative argument.
logbn = logan / logab
As long as we follow the pattern of the formula, you can make it workable on any base you choose.
given y = log2 3
can be rewritten as:
y = log10 3 / log10 2
How to we evaluate? We can either use our calculator and get y~ 1.585 . . .
Or, we can use logs. (No kidding, eh!)
y = log2 3
change to an exponential expression.
2^y = 3
then we apply the def of logs: log(2^y) = log3
y = log10(3) / log10(2)
Solving exponential equation.
Solve:
2(3^x) = 5
CAVEAT! (A latin word that means "warning" or "beware"):
log (x + 3) != (not equal to) logx + log3
instead, it becomes:
log(2(3^x)) = log5
log2 + log3^x = log5
log2 + x * log3 = log5
x * log3 = log5 - log2
x = log5 - log2 / log3 (at this point, we cannot continue without a calculator)
x = .8340437671
to check out answer: 2 * 3 ^ .8340437671 = 5
We just solved our very first exponential equation!
Let's solve another one:
19^(x - 5) = 3^(x+2)
(Note: Our reason for doing the log is to bring the exponents down.)
log 19 ^ (x - 5) = log 3^(x+2)
x * log19 - 5 * log19 = x * log3 + 2 * log 3
(x - 5) * log 19 = (x + 2) * log 3
x * log19 - 5 * log19 = x * log3 + 2 * log3
x * (log19 - log3) = 2 * log3 + 5 * log19
x = 2 * log3 + 5 * log19 / (log19 - log3) (now we use our calculator)
x ~ 9.166
log2 (x - 2) + log2(x) = log23
log2(x - 2)(x) = log23
x - 2(x) = 3 (we can remove the logs since they have the same base)
x ^ 2 - 2x - 3
(x + 1)(x - 3) = 0
x = -1, x = 3
STOP!
x = -1 doesn't solve the equation.
x = 3 works, though
log5(3x + 1) + log5(x - 3) = 3
log5[(3x+1)(x-3)] = 3
(use definition of logs:
5 ^ 3 = (3x + 1)(x - 3)
logab = n
a^n = b)
125 = 3x^2 - 8x -3
0 = 3x^2 - 8x - 128
(384: 16 and 24 as factors)
0 = 3x^2 - 24x + 16x - 128
0 = 3x(x-8) + 16(x-8)
0 = (3x + 16) (x - 8)
3x + 16 = 0, x - 8 = 0
x = -16/3 and x = 8
check for error:
- 16 /3 isn't a solution because after we plug it in, there is no value that 5 ^x that would be equal to a negative argument.
Thursday, November 5, 2009
Logarithmic LAWS!
So we started off the class with a little look at this website http://phet.colorado.edu/index.php. Take a look at it...Mr. Maks had a lot of fun on here, maybe you will too. It doesnt have anything to do with log functions, so dont get confused about that.
So here are the log laws that we learned today. I apologize that there are no slides to go along with the info, but it will be up on monday for sure.
1. log a (MN) = log a M + log a N
You might think that (MN) means multiply...but it DOESNT. It means you ADD the M and N.
2. log a (M/N) = log a M - log a N
Again...you might think this means divide, but it really means that you have to SUBTRACT.
3. log a M(exponent x) = (X)log a M
So for this rule you have to put the exponent number multiplied by the log a M.
All of these rules are on our formula sheet.
Next we did some examples to show how to do these new log rules. I'll try and write them out the best i can since i dont have the slides here.
ex) Write as a single logarithmic function: 2log x - log y - log z
First step: Use rule 3 to change the first part of the equation.
= log x(exponent 2) - log y - log z
Second step: Pull out the negative sign of the last part of the equation.
= log x(exponent 2) - (log y + log z)
Third step: Use rule 1 to combine the last part of the equation.
= log x(exponent 2) - (log (yz))
Last step: Since the equation is subtracting we can apply rule 2 to make the equation a single function.
= log(x(exponent 2)/yz)
It looks alot less complicated on the slides. The other examples will be up on monday. The assignment was to do Exercise 22 # 1-10 and Exercise 23 # 1-12
And remember...NO SCHOOL TOMORROW!
So here are the log laws that we learned today. I apologize that there are no slides to go along with the info, but it will be up on monday for sure.
1. log a (MN) = log a M + log a N
You might think that (MN) means multiply...but it DOESNT. It means you ADD the M and N.
2. log a (M/N) = log a M - log a N
Again...you might think this means divide, but it really means that you have to SUBTRACT.
3. log a M(exponent x) = (X)log a M
So for this rule you have to put the exponent number multiplied by the log a M.
All of these rules are on our formula sheet.
Next we did some examples to show how to do these new log rules. I'll try and write them out the best i can since i dont have the slides here.
ex) Write as a single logarithmic function: 2log x - log y - log z
First step: Use rule 3 to change the first part of the equation.
= log x(exponent 2) - log y - log z
Second step: Pull out the negative sign of the last part of the equation.
= log x(exponent 2) - (log y + log z)
Third step: Use rule 1 to combine the last part of the equation.
= log x(exponent 2) - (log (yz))
Last step: Since the equation is subtracting we can apply rule 2 to make the equation a single function.
= log(x(exponent 2)/yz)
It looks alot less complicated on the slides. The other examples will be up on monday. The assignment was to do Exercise 22 # 1-10 and Exercise 23 # 1-12
And remember...NO SCHOOL TOMORROW!
Wednesday, November 4, 2009
Logarithmic Functions
Today in class, we learned about Logarithmic Functions and how they are related to exponential functions. It looks a lot more confusing and difficult then it actually is, if that is consoling to any of you in any way. Remember that logarithms are exponents, and that not every logarithmic function has a solution. Homework check will probably be happening tomorrow or in the near future, so get exercises 12-20 done up. The assignment today was Exercise 21, questions 1-12.
Tuesday, November 3, 2009
Work Period
Today was early dismissal, so our math class was only 40 minutes long. Instead of learning something new, we were given time to work on our exercises that were previously assigned to us. There will be a homework check either tomorrow or Thursday on exercises 11-20, so make sure you get everything done! We should also be working on our Accelerated Math.
By the way, this week is Media Literacy Week!
Basically, its a week to promote the proper understanding and techniques of media education. You can learn more about it on this website: http://www.medialiteracyweek.ca
By the way, this week is Media Literacy Week!
Basically, its a week to promote the proper understanding and techniques of media education. You can learn more about it on this website: http://www.medialiteracyweek.ca
Monday, November 2, 2009
MOGA
Friday we didn't have math class because it was MOGA madness. MOGA is Most Outrageous Group Activity. On this day we all dress up for halloween and take bags of grain to the elevator to be taken to poverty-striken countries. You can buy a ticket for $2 and it gets doubled twice, so $8 is what is donated. Everyone uses motor-less transportation to get to the elevator and back. So yeah, that's fridays class.
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