CLEP Forum - CLEP Study - CLEP Testing - Study Guide and Strategies
 
  • CLEP Tips and Advice
  • CLEP Study Guide Information
  • Relieve CLEP Exam Anxiety
  •  

    Go Back   CLEP Forum - CLEP Study - CLEP Testing - Study Guide and Strategies > Main Category > Off Topic > What does this Flashcard mean or Do this math problem for me
    User Name
    Password
    Register FAQ Members List Calendar Search Today's Posts Mark Forums Read

    Reply
     
    LinkBack Thread Tools Search this Thread Display Modes
      #1 (permalink)  
    Old 07-23-2007, 06:05 PM
    studyhard studyhard is offline
    Viscount / Viscountess
     
    Join Date: Feb 2007
    Posts: 508
    Default College Mathematics

    Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them:


    If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ??????

    I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas?
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #2 (permalink)  
    Old 07-24-2007, 12:44 PM
    DixieGirl DixieGirl is offline
    Squire
     
    Join Date: Jul 2007
    Location: Tennessee
    Posts: 46
    Default

    Quote:
    Originally Posted by studyhard
    Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them:


    If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ??????

    I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas?

    Did you mean
    8^x = 15
    8^y = 25
    8^(2x+y) = ??

    I couldn't figure it out either, but I posted this on another forum and they said,

    "since 8x = 15 ... 82x = (8x)2 = 152.

    82x+y = 82x*8y = 152*25"

    Hope this helps!
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #3 (permalink)  
    Old 07-24-2007, 01:49 PM
    wannabeit wannabeit is offline
    Knight Champion
     
    Join Date: Aug 2006
    Location: LA
    Posts: 235
    Default

    Quote:
    Originally Posted by studyhard
    Aggh, I'm so fuzzy on this. This is a problem from the CLEP practice test. I'm putting parenthesis around things that are exponents, since there's no way to denote them:


    If 8(x) = 15 and 8(y) = 25, then 8(2x +y) = ??????

    I can't for the life of me remember how to work this problem and digging through my old math books shed no light. Any ideas?

    I think (x =15 and (y)=25, therefore 8 x (2 x 15+ 25)
    (30 +25) = 55
    8 x 55 = 440
    __________________
    MCSE (2003k), CCNP, Project+
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #4 (permalink)  
    Old 07-24-2007, 02:20 PM
    sgloer sgloer is offline
    Administrator
     
    Join Date: Feb 2006
    Posts: 725
    Default

    I believe this is how it works. There's a rule involving multiplying exponents with the same base number, so:
    8^(2x + y) is the same as
    8^(2x) + 8^(y) which is the same as
    8^x + 8^x + 8^y
    = 15 + 15 + 25
    = 55

    If you would provide the correct answer, that would help with reverse engineering the rule/method, because I might be remembering the rule incorrectly.
    __________________
    --
    Steve, Webmaster

    Last edited by sgloer : 07-24-2007 at 02:22 PM.
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #5 (permalink)  
    Old 07-27-2007, 02:45 PM
    tryingtesc tryingtesc is offline
    Minor Noble
     
    Join Date: Jul 2007
    Posts: 5
    Wink

    Ok...There are two ways to solve this problem...One is the use of Logs to find an unknown exponent. For example:
    8^x=15
    = log 15 base 8 = x (re-write in log form)
    =log 15 / log 8 (divide)
    =1.3022968652 (this is the "x" exponent)

    8^y=25
    =log 25 base 8 = y
    =log 25 / log 8
    = 1.54795206326 (this is the "y" exponent)

    Plug your now Known exponents into your equation:
    8^2x+y = 8^2(1.3022968652)+(1.54795206326) = 5624.9999 = 5625

    Second Way:
    They made this easy because of the same base 8.
    8^x=15 , 8^y=25, 8^2x+y=??
    8^2x+y = 8^2x * 8^y (re-write by seperating exponents)
    8^y is the easy part, we know that is 25
    8^2x is not the same of course as 8^x (which we know is 15) so what makes it different? Yes the "2". Well you squared the left side of the equation so you have to square the right side: 8^2x = 15^2. Now you know 8^2x = 225 (15 squared).
    So put the two together: 8^2x * 8^y is now the same as 225*25 = 5625

    ^ = raised to that exponent
    * = multiplication

    I hope this helps. It is a tough problem to explain on a message board.
    Best regards,
    Jason
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #6 (permalink)  
    Old 07-27-2007, 06:11 PM
    tryingtesc tryingtesc is offline
    Minor Noble
     
    Join Date: Jul 2007
    Posts: 5
    Default

    Quote:
    Originally Posted by sgloer
    I believe this is how it works. There's a rule involving multiplying exponents with the same base number, so:
    8^(2x + y) is the same as
    8^(2x) + 8^(y) which is the same as
    8^x + 8^x + 8^y
    = 15 + 15 + 25
    = 55

    If you would provide the correct answer, that would help with reverse engineering the rule/method, because I might be remembering the rule incorrectly.

    You have the problem set up nicely, however, you should multiply in rows 2, 3, and 4 instead of add. You can "split" the exponent, but you then must multiply. For example: If you had x^2+3; that would equal (x^2)(x^3), the bases are the same and in multiplication you would simply add the exponents making it x^5.
    You could not put them back together if you split them like this: x^2 + x^3. They would not be "like" terms, one is squared and the other cubed.

    I hope this helps with your reverse engineering because I am not familiar with that method.

    Best Regards,
    Jason

    Last edited by tryingtesc : 07-27-2007 at 06:15 PM.
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #7 (permalink)  
    Old 07-27-2007, 06:23 PM
    radyogyrl radyogyrl is offline
    Minor Noble
     
    Join Date: May 2007
    Posts: 6
    Default

    For this particular problem, you will square 15 which will give you 225 then mulitiply 225 by 25 which will give you 5625. I had a hard time figuring this one out when I was studying for the college math clep. I just played with it until I got the correct answer.
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
      #8 (permalink)  
    Old 07-27-2007, 07:14 PM
    sgloer sgloer is offline
    Administrator
     
    Join Date: Feb 2006
    Posts: 725
    Default

    Quote:
    Originally Posted by tryingtesc
    You have the problem set up nicely, however, you should multiply in rows 2, 3, and 4 instead of add. You can "split" the exponent, but you then must multiply. For example: If you had x^2+3; that would equal (x^2)(x^3), the bases are the same and in multiplication you would simply add the exponents making it x^5.
    You could not put them back together if you split them like this: x^2 + x^3. They would not be "like" terms, one is squared and the other cubed.

    I hope this helps with your reverse engineering because I am not familiar with that method.

    Best Regards,
    Jason

    Yeah, that looks right. Thanks Jason--I couldn't remember exactly how the rule went.
    __________________
    --
    Steve, Webmaster
    Digg this Post!Add Post to del.icio.usBookmark Post in TechnoratiFurl this Post!
    Reply With Quote
    Reply


    Thread Tools Search this Thread
    Search this Thread:

    Advanced Search
    Display Modes

    Posting Rules
    You may not post new threads
    You may not post replies
    You may not post attachments
    You may not edit your posts

    vB code is On
    Smilies are On
    [IMG] code is On
    HTML code is Off
    Trackbacks are On
    Pingbacks are On
    Refbacks are On



    All times are GMT -5. The time now is 07:27 PM.



    1 2 3 4 5 6 7 8