Results 1 to 8 of 8
Like Tree2Thanks
  • 1 Post By Plato
  • 1 Post By SworD

Math Help - Not looking for answer, just whining. For now.

  1. #1
    Newbie
    Joined
    Dec 2011
    From
    Central KY
    Posts
    6

    Unhappy Not looking for answer, just whining. For now.

    Not looking for answer, just whining. For now.-mathomir-page-28-problem-46.jpg

    I'm supposed 2 simplify this, and according to my Barron's Precalc The EZ way, the answer is B^n.
    So far, I've not gotten there, but hopefully I will soon. So far, I've not had issues with this problem set, but for some reason, this one and the next one (same type of problem) are stumping me.

    I will continue working with this and hopefully will get it, soon. I suspect that I need to start with the paranthetical cube in the numerator, but I feel like I'm 15 yrs old again (I'm in my 40s ☺) staring at my text book. Good grief, and this was my idea to try this book on for size.

    K
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,969
    Thanks
    1788
    Awards
    1

    Re: Not looking for answer, just whining. For now.

    Quote Originally Posted by kevro2000 View Post
    Click image for larger version. 

Name:	MathOMir Page 28 Problem 46.jpg 
Views:	2 
Size:	8.4 KB 
ID:	24662
    \frac{{{b^{{{\left( {2n + 1} \right)}^3}}}}}{{{b^n} \cdot {b^{4n + 1}}}} = \frac{{{b^{8{n^3} + 12{n^2} + 6n + 1}}}}{{{b^{5n + 1}}}} = {b^{8{n^3} + 12{n^2} + n}}
    Thanks from kevro2000
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Dec 2011
    From
    Central KY
    Posts
    6

    Re: Not looking for answer, just whining. For now.

    Plato, thank you: among my multi-solutions that I've come up with, your is the most logical, but not anything like the 2 or 3 that I came up with, and certainly not the "B^n" the Barron's EZ precalc gives as the answer. I'm going to study how to deal with exponents, and then come back to this section of questions in the book.

    Thanks, much,
    Kevin
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Member
    Joined
    Sep 2012
    From
    Planet Earth
    Posts
    215
    Thanks
    56

    Re: Not looking for answer, just whining. For now.

    Please note: a^(b^c) is NOT the same as (a^b)^c. Exponentiation is NOT associative.

    Are you sure that the initial equation isn't supposed to mean \left (b^{2n+1} \right )^3 instead? If it does, then the answer would indeed be b^n.

    \frac{\left (b^{2n+1} \right )^3}{b^n \cdot b^{4n+3}}

    Note that (a^b)^c = a^(b*c)

    = \frac{b^{6n+3}}{b^n \cdot b^{4n+3}}

    = \frac{b^{6n+3}}{b^{5n+3}}

    = b^{6n + 3 - 5n - 3} = b^n

    The fundamental property of exponents, valid everywhere, is that x^{a+b} = x^a \cdot x^b. So when b is negative, this leads to x^{a-b} = \frac{x^a}{x^b}. Make use of this property
    Last edited by SworD; September 6th 2012 at 07:29 PM.
    Thanks from kevro2000
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Newbie
    Joined
    Dec 2011
    From
    Central KY
    Posts
    6

    Re: Not looking for answer, just whining. For now.

    I'll double-check the book tonite, and make sure I've entered and am showing it correctly, as in the opening thread. I'm pretty sure I copied it and got it right, and i used MathOMir to turn it into an image: but just in case, I want to be sure. I understand both examples you two have given, and will look for more samples online to play with. My Barrons book only has two in this format (at least in this section of the book.)
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Newbie
    Joined
    Dec 2011
    From
    Central KY
    Posts
    6

    Re: Not looking for answer, just whining. For now.

    SworD, you are correct: I mis-keyed the equation. I see on my paper-work where I was first attempting it, I copied it correctly, but still worked it incorreclty. A few days later, on paper, I had copied it again, thinking I'd start over on it and do better, but I miswrote it down. From that, I used mathomir, and that is why the problem appears as such at the start of the thread.

    ► So, you are correct, the numerator is (b^2n+1)^3 All in parantheses, cubed. ◄
    Armed with the above comments, I will try this problme again, on my own. Thanks for the help, folks.
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Newbie
    Joined
    Dec 2011
    From
    Central KY
    Posts
    6

    Re: Not looking for answer, just whining. For now.

    Yay!!! Got it and the next (similar problem) in the book. Moving on to the word problems, now. Onward and forward...........I hope.
    Follow Math Help Forum on Facebook and Google+

  8. #8
    MHF Contributor

    Joined
    Apr 2005
    Posts
    16,453
    Thanks
    1868

    Re: Not looking for answer, just whining. For now.

    For me it's more often "Onward and aroundward"!
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 5
    Last Post: March 3rd 2013, 08:17 PM
  2. Converting my answer into the books answer
    Posted in the Algebra Forum
    Replies: 6
    Last Post: March 10th 2011, 03:06 PM
  3. Replies: 1
    Last Post: October 4th 2010, 05:46 PM
  4. checking my answer with another answer...
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: November 30th 2009, 08:12 AM

Search Tags


/mathhelpforum @mathhelpforum