Results 1 to 5 of 5

Math Help - solve inequality

  1. #1
    Member
    Joined
    Nov 2006
    Posts
    152

    solve inequality

    \log_{x}(\frac{5}{2}-\frac{1}{x})>1
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Joined
    Apr 2005
    Posts
    1,631
    Log(base x)[5/2 -1/x] > 1
    [5/2 -1/x] > x^1
    2.5 -1/x > x
    Multiply both sides by x,
    2.5x -1 > x^2 --------no change in sense of inequality since x is positive

    x^2 -2.5x +1 < 0
    2x^2 -5x +2 < 0
    (2x -1)(x -2) < 0

    x < 1/2 -----------(1)
    or x < 2 -----------(2)

    And, since there are no real logs of negative numbers, then,
    5/2 -1/x > 0
    5x -2 > 0
    x > 2/5
    x > 0.4 ------------(3)

    Investigate those three critical points.

    i) In the interval (0.4, 1/2)
    say x = 0.45,
    Log(base 0.45)[5/2 -1/(0.45)] >? 1
    0.2778 >? (0.45)^1
    No, so this interval is not a solution.

    ii) In the interval (1/2, 2)
    say x = 1.5,
    Log(base 1.5)[5/2 -1/(1.5)] >? 1
    1.8333 >? (1.5)^1
    Yes, so this interval is a solution.

    iii) In the interval (2, infinity)
    say x = 5,
    Log(base 5)[5/2 -1/5)] >? 1
    2.3 >? (5)^1
    No, so this interval is not a solution.

    Therefore, x = (1/2, 2), or 1/2 < x < 2 ------------answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by perash View Post
    \log_{x}(\frac{5}{2}-\frac{1}{x})>1
    Well, for starters to have valid argument for the logarithm we require
    \frac{5}{2} - \frac{1}{x} > 0

    \frac{1}{x} < \frac{5}{2}

    x > \frac{2}{5}
    so we must ensure that this condition holds.

    Now,
    log_{x}(\frac{5}{2}-\frac{1}{x})>1

    \frac{log(\frac{5}{2}-\frac{1}{x})}{log(x)} > 1
    by the change of base formula. (I'm using base 10 here, but feel free to use the base of your choice. It doesn't affect the answer.)

    Since log(x) is never negative:
    log(\frac{5}{2}-\frac{1}{x}) > log(x)

    Again, since log( ) is never negative, and the function 10^x is one to one:
    \frac{5}{2} - \frac{1}{x} > x

    Can you take this from here?

    -Dan
    Follow Math Help Forum on Facebook and Google+

  4. #4
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    First of all we have to put the conditions for the existance of logarithm and for the base:
    \displaystyle\frac{5}{2}-\frac{1}{x}>0 and x>0, \ x\neq 1\Rightarrow x\in\left(\frac{2}{5},\infty\right)-\{1\}
    If \displaystyle x\in\left(\frac{2}{5},1\right) then
    \displaystyle\frac{5}{2}-\frac{1}{x}<x\Rightarrow x\in\left(\frac{2}{5},\frac{1}{2}\right).
    If x\in(1,\infty)\Rightarrow\frac{5}{2}-\frac{1}{x}>x\Rightarrow x\in(1,2).
    Then x\in\left(\frac{2}{5},\frac{1}{2}\right)\cup(1,2)
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Forum Admin topsquark's Avatar
    Joined
    Jan 2006
    From
    Wellsville, NY
    Posts
    9,855
    Thanks
    321
    Awards
    1
    Quote Originally Posted by red_dog View Post
    First of all we have to put the conditions for the existance of logarithm and for the base:
    x\neq 1
    Ooh! I didn't think to look at that!

    (The story of my day, basically. )

    -Dan
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. [SOLVED] Solve the inequality 20-3y > y+4
    Posted in the Algebra Forum
    Replies: 4
    Last Post: September 19th 2011, 04:17 AM
  2. [SOLVED] Solve the inequality
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: August 27th 2011, 01:04 AM
  3. How do I solve this inequality?
    Posted in the Algebra Forum
    Replies: 7
    Last Post: December 24th 2010, 06:19 AM
  4. Solve the inequality
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: November 10th 2009, 07:09 AM
  5. Solve inequality help plz!!
    Posted in the Algebra Forum
    Replies: 1
    Last Post: September 10th 2008, 05:20 AM

Search Tags


/mathhelpforum @mathhelpforum