November 09, 2006

Reviewing For Logs and Exponents Test

At the bottom of this post are 4 questions for you to answer as you've done before with Jho-ahn's Bicycle and Why Not Zero?

But before we get to that, over the weekend you will be writing your first online quiz in preparation for the test. Here's what you have to do to get started:

The quiz will be written online here. It goes live at 3:30 pm tomorrow and will no longer be available as of midnight on Sunday. When you get to the site follow these instructions:

  1. Click on the big yellow "Sign Up" arrow.

  2. Use only your first name and last initial as indicated.

  3. Pick a username that will allow me to easily identify you, i.e. first name and last initial.

  4. Make up any password you like.

  5. Click on [Register] then [Search] by teacher's name (kuropatwa) and you'll find me.

  6. Click on the box next to Pre-Cal 40S and then [Register].

  7. Follow the instructions on the screen.

Actually, if you read each page carefully, you'll see that the sign up process is very straight forward and self explanatory. If you hit any snags email me and we'll sort it out together.

This quiz is timed. You'll only have 45 minutes to complete it once you've begun. It consists of 11 multiple choice questions. DON'T PANIC and I know you'll all do well. We'll talk more about this in class tomorrow. ;-)

Here are your practice questions for tonight ....

(1) Solve any one of the following equations for x. Show all of your work. Exact answers only, please. No decimal approximations. If you come to the party too late to solve a question then, like we did with Why Not Zero?, add a comment either agreeing or disagreeing with the solution(s) previously given. Begin your comment by saying: This is a comment on question A. (or B or C or D)

(a) 2x = 3(2x + 5)

(b) ex - 3e-x = 2

(c) log2x + log2(x - 3) = 2

(d) (logx)2 = log x2


  1. i'll start on question: (c)

    **note** log2 will be log(base of 2) to avoid confusion

    c)log2x + log2(x-3) = 2

    log2x + log2(x-3) = 2
    log2[x(x-3)] = 2
    log2[x^2 -3x] = 2
    2^2 = x^2 -3x
    *changed into exponential form
    4 = x^2 - 3x
    0 = x^2 - 3x - 4
    0 = (x+1)(x-4)

    x + 1 = 0 ; = - 4 = 0
    x = -1 ; x = 4

    check : - 1

    log2(-1) + log2(-1-3) = 2
    log2(-1) + log2(-4) = 2
    log2(-1 8 -4) = 2
    log2(5) = 2
    *since nothing equals 2 then -1 is rejected

    **ps** i'm not sure if i did this correctly i'd like someone to go over it for me.

    check: - 4

    log2(4) + log2(4-3) = 2
    log2(4) + log2(1) = 2
    2 + 0 = 2
    2 = 2

  2. Oh! All questions are difficult...
    Anyway, this is my solution for question D:

    (log X)^2 = log X^2

    (log X)(log X) = 2 (log X)

    (LHS)I get the factor
    (RHS) I use the power law

    (log X)(log X)/ (log X) = 2

    log X = 2

    Since this is common logarithm, our base is 10...

    log(base10) X = 2

    In exponenetial form

    10^2 = X

    X = 100

    To check:

    (log 100)^2 = 4

    2(log 100) = 4

    Hmmm..what do you think of my answer guys??
    Is it correct??

  3. Umm.. I'll start w/ question A.

    um, you log both sides..
    then the power law.
    distribute the 2x+5
    isolate the x's on the left side.
    factor out the x.
    divide both sides by log2-2log3

    ahh.. i'm not sure!! i tried (:

  4. This is a comment on question E:

    first: i will let "e" = log

    log^x - 3log^-x = log^2

    log^x - log^-x3 = log^2

    log^-x3/x = log^2

    -x^3/x = 2
    (multiple both sides by "x")

    -x^3 = 2x
    (divide both sides by "2" because we want "x" by it self)

    -x^3/2 = x
    (since the "x^3/2 is negative we must "flip it" to get rid off the negative sign)

    2/x^3 = x
    (multiple both side by "x^3")

    2 = x^4

    hmmmmmmmmm.....oh boy im not sure if my answer is correct but i tried to do this question many times and this is the best answer i came up with....if you have a better solution please teach me tomorrow or make a post :D

  5. This is a comment on question c because it was the easiest one for me to do=P
    So I basically got the same answer as Jefferson when I tried it because when you are adding logarithms with the same base you multiply. like what he did log2x + log2(x-3) = 2 would turn into this log2[x(x-3)] = 2. Then from there I did the same thing by changing it into exponential form and solving for x. it's hard for me to explain since i don't know the terms but all i know is that i got the same answer as jefferson with the answer being 4 and rejecting -1.

  6. re: ricardo

    i am so impressed on how you started off with question D!!! i never thought of seeing it as a difference of sqaures. i don't have the eyes for that. awesome job ricardo!

  7. Well... i'm so glad i remembered to check for the homework... i got kind of wrapped up in other things and was gonna go to bed... i really want to bring up my mark, but this is really all i got, so here goes...

    i tried to do all of these questions, but the only ones i could come up with an answer for were c, and d... which turned out were the answers already posted on the blog... c) x= -1 and x= 4 with the x= -1 being rejected because this number makes the equation in the brackets equal a negative number (which isn't suppose to happen)... and in d) i found that log x= 2 and that x= 100... i found c really easy, and when i saw that jefferson's answer was the same as mine, i was really happy that something actually sunk in... yay for the logs... and that d came out the same as ricardo's... which either means that i did it right, or we both did it wrong... either way i guess we find out tomorrow... sadly i just couldn't work out a or b, so i hope that i can pick up how to do them tomorrow... i'm pretty sleepy, but i'm going to try a and b again before class tomorrow to see if i get anything after some sleep... well good luck to everyone else and i'll see you in the morning! night!

  8. i helping on question (E)(for ruschev) that for some reason i cant find on the blog so i think it should have been (B) ...
    i'll start on question : (B)
    so think this was the original question : e^x - 3e^-x = 2
    so let begin solving for x

    so let e =log

    log^x -3log^-x = 2

    log^(x/-x^3) = 2

    we have to know that if log is by itself the its log10 so..

    x/-x^3= 10^2

    x/-x^3= 100

    multiple both sides by "x"

    (x) x/-x^3= 100(x)

    -x^3= 100x

    x^3 + 100x = 0

    i got suck here but i dont know if im right in doing this question but for the (C) + (D) i got the same answer as jefferson and ricardo but (A) + (B) in somewhat stuck in a pot hole but in did my best at it

    seeeyaa!!! and goodnight

  9. this is my solution for Question D
    I'm not sure with my solution...but everytime I check it..the two sides of the equation are equal..

    e^x -3e^(-x) = 2

    first I factor it out e^x

    e^x (1 - 3^(-1)) = 2

    divide both side by (1 - 3^(-1))

    e^x = 2/(1 - 3^(-1))

    e^x = 3

    In logarithmic form

    log(base e)3 = X

    It's the same thing as the

    In 3 = x

    That's my answer.. ln 3

    I'm not really sure but maybe there are other solutions..

  10. Oh that was Question B!!!

  11. hehehe....ooops....the question that i was doing was B! sorry guys

    thanks pat pat

  12. okay well , I guess IM a little too late for the party , the first two questiosn are tricky , and I did question C, I agree with jefferson's answers, I had done the same steps and I'm glad.