Results 1 to 5 of 5

Math Help - exponential inequations!

  1. #1
    Newbie
    Joined
    Feb 2010
    Posts
    13

    exponential inequations!

    2 inequations but please solve em without logarithm,..cuz we didnt learned logarithm...but if must do with logarithm go ahead i understand a bit of logs



    im sorry if i posted in bad section?really sorry..
    Last edited by icefirez; February 9th 2010 at 11:36 AM.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    -1
    e^(i*pi)'s Avatar
    Joined
    Feb 2009
    From
    West Midlands, England
    Posts
    3,053
    Thanks
    1
    Quote Originally Posted by icefirez View Post
    2 inequations but please solve em without logarithm,..cuz we didnt learned logarithm...but if must do with logarithm go ahead i understand a bit of logs



    im sorry if i posted in bad section?really sorry..
    Use
    Code:
    base^{exponent}
    for rendering exponentials in latex

    1. It follows that if \frac{1}{3^x+5} > \frac{1}{3^{x+1}-1} then 3^x+5 < 3^{x+1}-1. This is because when you divide by a fraction it's the same as multiplying by it's inverse. The smaller the denominator the larger the expression

    By the laws of exponents 3^{x+1} = 3 \cdot 3^x

    3^x+5 < 3 \cdot 3^x - 1

    Don't be scared by the exponent, you can treat it as normal

    6 < 2 \cdot 3^x

    3 < 3^x

    In most cases you'd then need to use logarithms but this one doesn't need them because if the bases are the same then the exponents must be the same:

    1 < x
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Feb 2010
    Posts
    13
    actually my proffesor said that's wrong...cuz i solved exactly like you first time...anyway thanks for trying...he said that we cant multiply with dominators cuz destroy inequality something like that anyway waiting for other responses
    by the way second inequation i solved...you dont have to solve again
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,713
    Thanks
    633
    Hello, icefirez!

    For the first one, your professor is right . . . up to a point.


    1)\;\;\frac{1}{3^x+5} \;>\;\frac{1}{3^{x+1} - 1}

    I'd like to cross-multiply, but I can't do it freely if either denominator is negative.

    We can see that (3^x+5) is always positive.

    But (3^{x+1} - 1) could be negative.
    When is it positive?

    . . 3^{x+1}-1 \:>\:0 \quad\Rightarrow\quad 3^{x+1} \:>\:1 \quad\Rightarrow\quad 3^{x+1}\:>\:3^0 \quad\Rightarrow\quad x+1 \:>\:0 \quad\Rightarrow\quad x \:>\:-1

    So, as long as {\color{blue}x \,>\,-1}. both denominators are positive. .[1]


    Cross-multiply: . 3^{x+1}-1 \;>\;3^x+5 \quad\Rightarrow\quad 3^{x-1} - 3^x \:>\:6

    Factor: . 3^x(3-1) \:>\:6 \quad\Rightarrow\quad 3^x\cdot 2 \:>\:6 \quad\Rightarrow\quad 3^x \:>\:3 \quad\Rightarrow\quad x \:>\:1

    This satisfies [1], so the soluton is: . \boxed{x \:>\:1}

    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor

    Joined
    Aug 2006
    Posts
    18,617
    Thanks
    1581
    Awards
    1
    And Soroban is correct ‘up to a point’!
    Note that 3^{x+1}-1<0 for all x<-1.
    Because 3^x+5>0 for all x that means
    \frac{1}{3^x+5}>0>\frac{1}{3^{x+1}-1} for x<-1
    So the solution \boxed{( - \infty , - 1) \cup (1,\infty )}
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 2
    Last Post: June 9th 2011, 09:54 AM
  2. Question about inequations
    Posted in the Algebra Forum
    Replies: 2
    Last Post: July 31st 2010, 01:06 AM
  3. Inequations
    Posted in the Algebra Forum
    Replies: 1
    Last Post: March 31st 2010, 12:16 PM
  4. Inequations
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: March 27th 2010, 06:10 AM
  5. solving inequations
    Posted in the Algebra Forum
    Replies: 6
    Last Post: December 15th 2008, 08:27 AM

Search Tags


/mathhelpforum @mathhelpforum