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Reviewing my logarithms, need help solving this inequality

  1. Dec 26, 2012 #1
    1. The problem statement, all variables and given/known data
    [itex]\frac{2^{x+1}-3}{2^{x}-4}\leq1[/itex]

    2. Relevant equations



    3. The attempt at a solution
    When I go through it, I keep getting [itex]2^{x}\leq-1[/itex]

    I don't think the answer is suppose to be complex.... let me show you my work in a file.

    It's extremely embarrassing that I don't remember this. Makes me feel pretty stupid....
     

    Attached Files:

  2. jcsd
  3. Dec 26, 2012 #2
    The first step is problematic, because 2^x - 4 could be negative. Multiplying both sides by this factor would then result in the inequality flipping. You should be multiplying both sides by (2^x - 4)^2 instead, which will be a positive quantity.
     
  4. Dec 26, 2012 #3
    I'm not sure that I follow. I plugged it into wolfram, and it comes up with a complex solution. If the make it less than, it is real.


    Even if the sign flipped, it still can't work out to be real.
     
  5. Dec 26, 2012 #4

    Ray Vickson

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    Just set ##z = 2^x## and find where ##(2z - 3)/(z-4) \leq 1.## If you can satisfy that with some ##z > 0## you will get a real ##x = \log_{2}(z) = \log(z) / \log(2).##
     
  6. Dec 26, 2012 #5
    To remove the fractional term in
    [tex]\frac{2^{x+1}-3}{2^{x}-4}\leq1[/tex]
    You must multiply both sides by the denominator squared and not just the denominator because it need not be positive. So, we get
    [tex](2^{x+1}-3)(2^{x}-4)\leq (2^{x}-4)^2[/tex]
    instead of whatever you wrote for your second step.
    Expand out the terms, collect them on one side, and refactorise.
     
  7. Dec 26, 2012 #6

    haruspex

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    Isn't it simpler just to treat the two cases separately? The positive sign led to a contradiction, so we can discard that.
    No, I think that works. Try it again. If still stuck, please post your working for that case.
     
  8. Dec 26, 2012 #7

    HallsofIvy

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    Alternatively, do it as two separate problems.
    1) if [itex]2^x- 4\ge 0[/itex], multiplying both sides by it gives
    [itex]2^{x+1}- 3\le 2^x- 4[/itex]
    Of course, [itex]2^{x+ 1}= 2(2^x)[/itex] so that inequality is the same as [itex]2(2^x)- 3\le 2^x- 4[/itex]. Subtracting [itex]2^x[/itex] from each side and adding 3 to both sides we get [itex]2^x\le -1[/itex] which is, as you say, impossible for x a real number.

    2) If [itex]2^x- 4< 0[/itex], multiplying both sides of it changes the direction of the inequality: [itex]2^{x+1}- 3\ge 2^x- 4[/itex] which is the same as [itex]2(2^x)- 3\ge 2^x- 4[/itex]. Now, doing the same as before gives [itex]2^x\ge -1[/itex] which is true for all x. Since this was under the condition that [itex]2^x- 4< 0[/itex], the inequality is true for all x satisfying [itex]2^x< 4[/itex] which is the same as saying that x< 2.
     
  9. Dec 26, 2012 #8

    SammyS

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    You are correct that (2x - 4) is positive for some values of x and negative for others. If Astrum multiplies both sides of the inequality by (2x - 4), then he/she should look at 2 cases separately:
    1) 2x > 4, i.e. x>2 .

    2) 2x < 4, i.e. x<2 .​

    Alternatively, your suggestion of multiplying by (2x - 4)2 eliminates the problem of looking at two cases separately.

    Another method, one that's usually covered in College Algebra courses, is as follows:

    Subtract 1 from both sides, then use a common denominator to get a single rational expression on the left. That results in a rational expression being compared to zero. For the problem at hand, this resulting rational expression is fairly simple because 2x+1 = 2∙2x .
     
  10. Dec 26, 2012 #9
    Ah, Alright. I see. The [itex]\frac{1}{2^{x}-4}[/itex] was where I messed up.

    It's been a while since I've seen these problems, I guess some memory lapse is to be expected.
     
  11. Dec 27, 2012 #10

    SammyS

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    What did you get for a solution ?
     
  12. Dec 27, 2012 #11
    x<2

    the original problem was wrong, if it meant "equal to" also, because it would lead to a complex answer.
     
  13. Dec 27, 2012 #12

    haruspex

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    That doesn't make the problem wrong. It asked you to find those x that satisfied the inequality, and you found them.
     
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