(1) Solve for x:
(a) log8x = 2
(b) log4x = -(3/2)
(c) logx27 = (3/2)
(d) logx5 = -(1/2)
(3) Given that log109 = 0.95 and log102 = 0.30, without using a calculator evaluate:
scroll way down for the answers ....
don't look until you're tried all the questions ....
HEY .... no peeking!!!!!! ;-)
Wonder what this picture has to do with logarithms ....
... something ... can you figure out what? ;-)
(1) (a) 64 (b) 1/8 (c) 9 (d) 1/25
(2) (a) 5log2M + 4log2N (b) you figure it out. ;-)
(3) (a) 1.9 (b) 0.475 (c) 0.65 (d) -3.30
October 31, 2006
(1) Solve for x:
We were given time to solve this equations, after the time was done we explained how we got our answers.
To find the area of triangle ABC we had to find the points of A, B ,C is this is the step by step solution.
Point A is found by plugging our x value into the equation y=3x, our given value for x was zero(0), and any power to the exponent of 0 will always equal to 1, which gives us the coordinates (0,1) and that is the coordinate for point A
Point B is found by plugging in our x value also into the equation, we got the value of x by the ordered pair below point B which was (2,0), when we plug in the value of x into the equation y=9, so the coordinates of point B would be (2,9)
Point C is found because we are given the x value which is also seen by above in Point B, and the point Below it, because there on the same x distance, but just different y distance,so the x value is 2, and the y distance is going to be the same as point A because it lies on the same height as point A, so the y value is 1, so the coordinates of Point C is (2,1)
So the solution for question #1 is 8 units squared
the solution to question # 2 is found by finding the value of base of x, we simplified it into becoming 2³(2x-1)³=5³, it had the same exponents so we dropped them and simplified it to [2 (2x-1)] ³ = 5³, then we simplified it by it equaling 4x-2 =5, we then moved the -2 onto the other side od the equal sign and we get 4x=7, we then isolate the x and get x= 7/4
So the solution to question # 2 is x=7/4
It was a homework question for two classes ago. First off we went by re-writing the question into making it look like this 2^x/ 2 - 2^ x= 2^-3, to make the fraction disappear in the equation we multiply by 2 on both sides of the equal sign, therefore the equation will now look liek this 2^x - (2)2^x=(2)2^-3, at this point dont add because you'll end up getting the original equation, so what you do now is solve by factoring. You will isolate the varaible 2^x. So it will be written like this 2^x(1-2)=2^-2, it will then end up like -2^x = 2^-2 , after simplifying. Therefore we concluded there is no solution because even if you were to multiply everything by 1 youll still have a postive power one side, and a negative power on the other, therefore staing its not true.
So the solution for question # 3 is NO SOLUTION
When we were asked to simplify (3^ 5)(3^7), we can say that when we are multiplying powers with the same base we add the exponents, therefore it can be simplified by writing it like this 3^(5 +7), in other words 3^12
So the solution for Question # 4 is 3^12.
Question # 5
when we are asked to simplify 5^11/5^4 we can say that when dividing powers with the same base we subtract exponents. If you apply this concept you will get an outcome like this 5^(11-4) or in other words 5^ 7.
So the solution for question #5 is 5^7
Question # 6
When we are asked to simplify (4^3)^5, we can say that when a power is raised to a power we multiply the exponents, so it would look like this [4^(3)(5)] or in other words 4^ 15
So the solution to Question#6 is 4^15
Question 7-9 remember LOGARITHMS ARE EXPONENTS!! BECAUSE MR.K KNOWS THAT AT LEAST 6 OF US ARE GOING TO FORGET. These questions are also for us to adapt on knowing how to run our tools backwards, and fowards.
When asked to simplify log2^8 + log2^32, we look at as its adding expoents, so your multiplying powers that have the same bases, so that question can be simplified as log2^(8)(32) or log2^256.
So the solution for Question #7 is Log 2^256
When asked to simplify log3^81 - log3^9, we look at it as we are subtracting exponents meaning we are diving powers. Thefore the equation can be written like log3^(81/9) or in other words log3^9.
So the solution for Question # 8 is log 3^9
When asked to simplify 2log5^125, we see here that we are going to multiply powers, which can be also known as to raising a power to a power. So it can be written like log5^125^2,or in other words log 5^15625
So the solution for question #9 is log 5^15625.
log a^(mn) = log a ^m + log a^n
log a ^(m/n)= log a ^m - log a^ n
log a^(m^c) = C loga^m
The above was written on the whiteboard after the qesutions were completed. These are the concepts we applied to our questions from earlier to get our solutions.
We were then given some new questions, and we were asked to write them as a single logarithm
The four questions were
a) 1/2 log a^9 - log a^3
simplify the equation as log a^3 - log a^3 which will give us loga ^ 3/3 by the corresponding concept of subtracting exponents meaning your dividing powers, you will then get to simplify as loga^1 which will give us 0.
The solution to a) is zero(0)
b)log a^6 + log a^4
adding exponents is the same as multiplying powers, so therefore it would simplify as loga^(6)(4), which in other words can be loga^24.
The solution to b) is loga^24
c) 2log a^5 + loga^4
simplified as loga^25 + loga^4, then we apply the adding exponents meaning multiplying powers so we simplify it by writing it as log a^ 100.
The solution to c) is loga^100
d) log a^ 6 + log a^ 8 - 1/2log a^16
therefore we simplify the adding of exponents as multiplying powers, and then simplifying the multiplying of exponents as power raised to a power, so it can be rewritten like,loga^(48)-log a^(16^1/2), which then can be simplified as loga^ (48/4), which can even be more simplified as loga^12.
The solution to d) is log a^12
GET OUT YOUR MATH DICTONARIES
NEW PAGE EVERYBODY
#1 the logarithm function is the inverse of the exponental function.
#2 the logarithm function is the function that turns powers into exponents.
REMEMBER: A LOGARITHM IS AN EXPONENT!
And thats all for today folks.....and the next scribe is Richard..he knows already =)
October 30, 2006
We were given questions as usual:
a) f(x) = sinx on the interval [-pi/2, pi/2]
*sketch a normal sine function to start.
since there was a domain given erase the graph so that the sine function is between -[pi/2 and pi/2]
a) g(x) = cosx on the interval [0, pi]
*pretty much the same as the top
c)h(x) = tanx on the interval [-pi/2, pi/2]
*and again the same steps apply here
next we were asked to sketch the inverse of the three graphs
*draw the line of reflection y=x.
the coordinates change from (x,y) to (y,x) ei (2,3) to (3,2)
the original graph is just flipped over the line of reflection
*same rules apply
c) *this one may be a little tricky
draw the line of reflection to start out with
since the graph is being reflected, so are the asymptotes
the asymptotes are now at x = 0 (orange)
Next we were asked to sketch:
y = 2^x
*the two most easy points you can get are when you input 0 and 1. If you replace x with 0 as an input you get an output (y coord) as 1. if you replace x with 1 as an input, then you get an output of 2
y = 3^x
next we had to sketch the inverse of the two :
*draw the line of reflection first. since the graph is being flipped over the line y = x, the asymptote(blue) is also being flipped.
*the new asymptote is at y = 0 (seen in red)
*if you forget the order of pairs (x,y) are flipped to (y,x)
y = 1/(2^x)
y = 1/(3^x)
when you input a number you nonetheless gt an output of the result. the inputs of these graphs are the exponents(x). And from the input you get an output of a power. IE 2^2(input/exponent) = 4(output/power)
Now when you do the inverse, the input is the power and the output is the exponent. it's basically saying it backwards, that's pretty much what an inverse is...sorta.
NOW it is time for me to bring in the logarithm aka the LOG(no not the tree log) . a log is a number but not any number. its a number that is formed from a result. An example Will come up later on this scribe post.
b = base
a = exponent
c = power
this is the basic formula for powers and exponents
a logarithm reverses that equation:
logb(c) = a
this equation says that a logarithm is an EXPONENT!
you WILL i repeat WILL hear Mr k repeat this 100 times! because students will often forget about this
here's an example:
23 = 8
log2(8) = 3
*the 8 and 3 switch because a logarithm is an exponent.
53 = 125
the next scribe will be.....jhoann
October 27, 2006
The morning was nothing new since we had a test the entire period, so there was one or two new bits of information during the afternoon class.
When we first came in the usual questions were on the board.
Solve for x:
There was one last question which gave people plenty of problems since almost everyone was apparently making the same mistake in the solving.
LET a = 2x
a=1 or a=2
The common mistake that people were making was doing something along these lines:
For the next step they would divide both sides by 2^x but did not cancel the -2 at the end of the equation. This cannot work and was explained to us why in this example.
----- = 7
If we were to follow the idea above then only the 3 would cancel and the answer would be 7 instead of the proper answer being 10/3
We also learned about some other minor bits of information but it's not really important as it was more about some of the graphing calculator functions.
The next scribe on Monday will be Jefferson.
October 26, 2006
Hi it's Ashley and I'm the scribe for today!
In class Mr.K wrote down 5 numbers on the board and told us to come up with 4 different ways to write down the number using exponents. The numbers were 2, 4, 3, 1/2 and 4/9. Seems hard huh? But don't worry once you see the pattern it gets easier.
By the way for those that don't know, this symbol ^ means to the power of. So for example 2^4 means 2 to the power of 4 which equals 16.
2^x=32 (well remember 32 can be written as 2^5)
x-1 = 3
In these questions like the questions above you are asked to prove and find the value of x. The key to these questions is that the idea is to find another way to write the number using a useful base. What you want is to look for a common base for both expressions. The reason why you look for a common base is because if the bases are the same then the exponents must equal the same. Then you just solve for x. It will be further understood when you see the questions below since I'm poor at explaining -_- Oh but the questions might look a little weird since it's hard to write exponents on the computer so here's an example of how it looks like on paper:
1) 2^3-2x = 1/4^x-3/2
the next step is to find another way to write the number using a useful base. in this case we want the base 1/4 to be 2.
2^3-2x = (2^-1/2)^x-3/2 (1/4 equals 2^-1/2)
2^3-2x = 2^(-x/2+3/4) (what you do here is multiply -1/2 to x-3/2 which gives you -x/2+3/4)
now that you have the same base for both expressions you can now solve for x.
3-2x = -x/2+3/4
4(3-2x) = (-x/2+3/4)4 (you want to get rid of the denominator which is 4 so you multipy both sides by 4)
12-8x = -2x+3
12-3 = -2+8
9 = 6x
3/2 = x
3^4x-1 = (3^3)^2x (27 equals 3^3)
3^4x-1 = 3^6x
now sove for x
4x-1 = 6x
-1 = 2x
-1/2 = x
3) (3^x)(27) = 81^2x+1
now look for another way to write the numbers using a useful base. for this question a useful base would be 3 since both 27 and 81 are multiples of 3.
(3^x)(3^3) = (3^4)^2x+1 (27 equals 3^3 and 81 equals 3^4)
*now remember when you multiply powers with the same base you add
3^x+3 = 3^8x+4 (3^x*3^3 the power x and 3 are added together so they become x+3)
solve for x:
x+3 = 8x+4
3-4 = 8x-x
-1 = 7x
-1/7 = x
like in previous questions rewrite the base so that both numbers have the same base. at first some would think of changing the base 4 into base 1 by having the power be 0 but then the 0 would cancel the whole exponent. so instead you should do this:
4^2x-4 = 4^0 (the 4^0 would equal 1 since any number to the power of 0 expect 0 itself equals 1)
then you go on from there solving x
2x-4 = 0
x = 4/2
x = 2
For this question at first it looks hard but actually what you do is balance the equation first so you can get the power by itself like this:
5^(2x-1)+2 = 3
5^(2x-1) = 3-2
5^(2x-1) = 1
next step is similar to the previous question in that you re-write the 1 to be the same base as 5 by having the power be 0.
5^(2x-1) = 5^0
2x-1 = 0
2x = 1
x = 1/2
So this is the end of my post and if there are any mistakes just comment. Also if it's hard to read sorry about that but for some reason my spacings didn't turn out right. I wonder why it does that? By the way the homework is Ex.20. Apparently if you understand what we did in class the homework will be super easy but remember we still have the test on Trigonometric Identities tomorrow guys so study hard!! Mr. K said the last page of the test will have the identities on it so that's a relief. Good luck everybody!
Okay dokes, this is my blogging on blogging, and I would like to say the trigonometric identities unit was great, I thought it was a lot of fun. I wasn't the best at it but I mean, I just like those kinds of things. I got it at times, and I had some questions at times, but I believe I will do okay tomorrow for the test, and I hope EVERYONE DOES GOOD TOMORROW ! don't forget to study guys ! that is all, bye bye.
Hopefully I'm not too late to put in my BOB. I have to admit, trig identities is a tough unit. In the terms of Mr. K, I think I would be better prepared if I had just taken the chance and fall off my bike a couple more times. It's not like I've given up on it, I know I could've studied more to have a little more knowledge to take along with me to test day. Looks like I'm hitting the books. Good luck to everyone.
Posted by JessicaJill at 8:47 PM
Trig identities was an OK topic for me for me. I especially like the double angle identities and i don't know why. Fellow classmates think I'm weird cause they say its harder. Though how Mr. K got the sin and cosine stuck in our heads by the weird dance, i think it was pretty clever, because i have it stuck in my head. Ha ha and to tell you did we ever LOOK WEIRD!! I probably like this unit better than the rest. I guess because i was exposed to it more than the other units. The test is tomorrow and i hope i do better than average. Like the bicycle speech we had, i wanna RULE. Not only in this unit but in every other unit as well. I hope everyone does well for tomorrows unit test.
The subject for this topic was pretty OK , i like hate it when you have to prove identities , they are not easy. Tell you the true i really don't like this unit. But i still trying. Just need more practices. How you explain it was very good , i understand it more this year than last year.Well i hope i do good on the test ,but i have a feeling that I'm not going to get a high mark. I guess the thing about identities is that you have lots of ways to do it not just one. So, what you think you got wrong , actually is right.other wise yeah. Bye
Oh boy, i hope this isn't too late... i was told today that the test is on friday, which means i still have some time to finish this... i wrote BOB on my hand but didn't see it until now... well this unit seemed pretty easy in the beginning... i found the first equations super simple to solve and follow... but as we got harder and harder my obvious troubles with equations seemed to creep up again... and when i lose faith in my knowledge, i lose all my composure and can't solve a thing... even if i do know my stuff deep down... my math expertise has really fizzled and now i need to pay attention more then ever... i think the hardest part for me is putting it all together... for example i may know my stuff, but when its phrased differently then i'm use to i can't seem to figure it out even if its the exact same question... like the pre-test my brain just said "i don't know what its talking about" and i couldn't figure it out... but when it was explained i was like "oh hey that's just like this!!".....i think i really need to just keep getting up on that bicycle and trying again no matter how many times i fall.... i'm doing my best, but my best isn't good enough so i have to try harder... and i will.... i guess that's all for now i gotta be getting to bed.... Night all!!!
October 25, 2006
i think i'm going ok on trigonometric identites although the pre-test was hard. for me anyways because sometimes i don't understand the question and sometimes i don't know how to solve it but i'm understanding now. i think. i just need to practice lots especially on the proving part because i get stuck a lot and i need to start recognizing the identites too but i sort of understand double angles now. so the test is on friday. good luck guys!
Identites...I'm pretty sure I understand the concept. The only thing that's giving me somewhat a hard time is memorizing the double angle identities. The pre-test for me was pretty straight forward until the last question where we had to prove the identity because I didn't know what could be substituted in for the cos2θ. My pre-test group did an awsome job though...good luck everyone =)
I have to say that this unit was confusing at first. When I learned it, it was all right, I understood it. Then there were rules like the sine/cosine dance we did, as well as several double angle identies. There could be so many different ways to solve an equation and sometimes it was hard for me to figure out how to do it. I can memorize the identities, but I have a bit of a hard time applying them to different problems. This is really similar to algebra to me, but with more variables. Until the test, I'm going to try and focus on getting help and doing some of the problems I don't know.
Complete the table of values:
October 24, 2006
Pre-test tomorrow... I don't know if I am ready or not, I know how to apply everything, it's just i can't identify it right. I know this unit gave me a hard time last year also. Looking forward to the next unit :P I'll finish up my little BoB here.
Good luck tomorrow everyone!
So at first, i understand this unit very well, and i think it was pretty much easier than the other units, but lately, i haven't been to class so i missed out some lessons and i feel like i get really behind, but hopefully, in two days before the test, ill understand and know what to do, especially with those tricky questions!
Anyways, wish all of you guys will ace the test :)
I guess this is the easiest unit so far. Maybe because I like Algebra and it's fun to untangle things out. It's just like a mystery that you need to solve using a lot of proven data. Aside from that, there is a lot of possible ways to use in order to get the answer. You just have to think wide. Think of the other possibilities.
I don't have any difficulties regarding this unit. It is just some of the questions are tricky. You can get it simply in a short way. But if you miss some ideas, it may lead you to a long and tangled equation.
I hope we all do good on our test. Do your best! Don't ever, ever give up!
identities, identities, identit..... well it's a good topic to do but i still have some problem on solving them if i start the question out right, later i will get confuse on what do next or which steps would get me there to get the answer. only if had enough time to solve it i may get the answer but practice, practice make some people better at these things ummm..... well goodluck to all and let the good come to you wuhahaha!!!!
Simplify: sin2θ/sinθ, now from here, automatically you can go straight to 2cosθ, which is the answer. But just to show you how I got it, then here it is, because math, is, FUN!
For question number 3, this was how it was solve:
The lesson about identities was easier for me than the past lesson. I found it very interesting because when we first took our exercise, I answered it correctly. It helped me to understand it better when I used to answer the questions in the assignments and it amazed me because most of the questions given in the class are the ones given in the assignment. However, there are times that even though I did the assignment I encountered some difficulties in answering some seatworks but then I asked some of my classmates to help me solved the problem and I thank them for doing such a wonderful thing. On the other hand, Mr. K did explained to us everything in a fully detailed manner, so it also gave the whole class better points to remember in solving identities.
In general, the lesson about identities was a challenge for me because it summarizes everything we took starting from lesson one up to the present lesson. Therefore, there is a need for me and for the others to study more because this can tell whether or not we understood the whole lesson.
October 23, 2006
New unit and at the very least I'm having an easier time with the work then previous questions, though not by much. During homework, or the quiz today for example I end up making all sorts of tiny mistakes that end up costing me the question. If anything I am going to need practice and plenty of it if I even hope to get good marks for this semester. Overall I'd probably just need some basic explanations and practice.
third bob, wow. hm, the topic identities. so far, i think it's okay. like i know what to do, yet i get stuck because i can't find a way to solve it. but mr. k said that we had to find another point of view to find the answer and there are more than one way of solving the identity. and that's pretty much all i need to focus on, but everything so far - i'm alright with (: goodluck guys.
Well so far so good for me. Proving identities aren't that difficult to do. However doing questions similar from the practise quiz that we had today was a little hard. I really had to think. I'm ok with solving difference identities. We didn't cover the tangent differences identities yet though. but I'm sure I'll do fine with them if we go over that tomorrow. Overall I think I'll do fine on the test. jut some more practise and review and I'll do well =)
Hey you guys, it's JessicaJill, and I was your scribe for today's double class. In the morning we had a five question quiz to do in 5 minutes. Here it is.
1. secΘ - tanΘ is identical to:
a) cosΘ/1 + sinΘ
b) sinΘ/1 + cosΘ
e) None of these
SOLUTION: The answer is a. Here's why:
d) cosΘ/1 + sinΘ
e) sinΘ/1 - cosΘ
SOLUTION: The answer is c.
4. Given than cosΘ = -1/3 and sinΘ is greater than 0, then sinΘ equals
SOLUTION: The answer is a.
5. Given that cscΘ = -4, Θ is greater than π and less than 3π/2, then tanΘ equals
SOLUTION: The answer is e.
Because a majority of the class had a little difficulty with these questions, it wasn't worth marks.
Then we were given three questions on the board to solve:
After that, he asked us what quadrant cos2α was in from question 3, and we all had different answers. A lot of us were a little afraid of being wrong, so he put us in groups to think about what quadrant we thought it was in. With several different answers given, we came to the conclusion that it was in quadrant III. If you think about what our answer was, where's it's negative and where the related angle is, the answer is clear.
In our afternoon class we got some notes to put into our math dictionaries.
We then were given 5 questions on the board, all that were on (a) previous exam(s), but didn't have enough time to go through the answers in class, so we were told to do it for homework.
1. Solve for θER in radians to 3 decimal places [calculator allowed]:
4cos(2θ) + 3 = 0
2. Simplify tanθ/cscθ
3. α and β are quadrant II angles, sinα = 1/3 and sinβ = 2/3. Find the exact value of cos(α + β).
4.Write and equivalent expression: cos(3x)cos(2x) + sin(3x)sin(2x)
5. Prove: (1/1 - sinθ) + (1/1 + sinθ) = 2tan²θ + 2
Well, that's it. Sorry it took me so long to get it up. I had a lot of work in the last week and I wanted everything to be perfect. This post is still a little incomplete, but I wanted it to be published already.
October 20, 2006
Hey people! This is Jessicca_198. Today, all we really did was Exercise 18, so there isn't really much for me to inform everyone on. I'm not exactly sure this counts as a scribe post, but the next person to be scribe after me is JessicaJill. Sorry I couln't have anything more for you guys. See you all on Monday!
Posted by 10101924 at 7:40 PM
October 19, 2006
a. sin 100° cos 50°+ cos 100° sin 50°
we know that this is the sine law
sin (100° + 50°)
b. cos π/3 cos π/12 + sin π/3 sin π/12
from our "dance" we know that this is the cosine law
cos (π/3- cos π/12)
and we also know that the sign changes
we then used a common denominator
cos (4π/12- π/12)
and we get (3π/12) we reduce it as cos (π/4)
which has a value of
d. cos(π+x)= cosx
cosπ cosx+ sinπsinx
-1 cosx -(0) sinx
PROOFS OF THE SUM AND DIFFERENCE IDENTITIES
October 18, 2006
Hello everyone. It's me Gerald again and this is my second scribe post. When we entered class, Mr. K put questions on the board for us to do.
SOLUTION IN RED
cos (α+β)= cosα cosβ + sinα sinβ
thats all for today. Its my nap time anyway =) next scribe is jaifel. See you next time
October 17, 2006
Good evening fellow students, names Steve =o. I hope you guys enjoyed the formulas and the 'plug and solve' methods on previous units because it will be rarely used in this unit, Identities. Like Mr.K mentioned, practice is what will be the key to you mastering these problems, so the beginning of the first class of todays two-class day he threw at us these problems:
Fear my fish/Alpha variables =P
Well, goody. Those were the simplification problems, now that they're done with, we have some questions where we have to prove that one operation equals the other. Im kind of running behind schedual, those pictures were hard to put together =P, thats was fun, but lets see if we can deal with simple text typed ones.
If starting with the Right side:
If starting with the Right side:
Well, that was the end of the first class. It was hard to get used to, so usually alot of explaination was needed to help understand Identities more, which is why we only had 6 questions. Great! you got this far, it's intermission time =P. It was lunch for us, go get a snack or something and come back.
Welcome back! Being seated we noticed Mr.K had put up more questions for our... pleasure? Well, lets get to it! Noted, he has raised the bar once more, these were much tougher for us, noticably.
3)[1/(1+sinx)] + [1/(1-sinx)] = 2sec2x