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Determining the form of the real solution set?

  1. Dec 9, 2015 #1
    1. The problem statement, all variables and given/known data
    ##\frac{log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x)}{log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} \geqslant 0 ##
    The set of all real solutions to this inequality is of the form:
    ##(a) ## ##(a,b) \cup (b,c) ##, ##(b) ## ##(-\infty,a) \cup (c,\infty) ##, ##(c) ## ##(a,b) ##
    for some real numbers ##a,b,c ## such that ##-\infty<a<b<c<+\infty ##
    2. Relevant equations
    3. The attempt at a solution
    First i found the points at which the numerator and the denominator equal ##0 ##. I found that from the following:
    ##log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x) = 0 ##
    ##log_{2x^2 + 2x + 3}(x^2 - 2x) = 1 ##
    ##2x^2 + 2x + 3 = x^2 - 2x ##
    from this quadratic equation i find that the zeros of the function are ##x_1=-3 ## and ##x_2=-1 ## and let them be ##a=-1 ## and ##b=-3 ## for now. i also found from the plot of the given quadratic equation that ##(-\infty,-3)\cup(-1,\infty) ## makes function a positiv one and the values in between ##(-3,-1) ## make it negative. I used that later for the rational number chart.
    Similarly ##log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10) = 0 ##
    ##x^2 + 6x + 10 = 1 ##
    ##x = -3 ##. This function is positive for every other value.
    Now using the chart of rational numbers, determining when the whole ratio is positive and when is negative i get the positive values to range from
    ##(-\infty,-3) \cup (-1,\infty)## which then makes my answer of the form ##(-\infty,b) \cup (a, +\infty)##.
    First of all, its not given, second, its wrong. The solution is i know the ##(a) ## answer. I just dont know how to get to it. Could you help?
     
    Last edited by a moderator: Dec 9, 2015
  2. jcsd
  3. Dec 9, 2015 #2

    RUber

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    It looks like you have done well to eliminate the values which would drive a division by zero. However, there are other considerations you must apply to satisfy the domain for the logarithms.
    First, your base may not be zero. ##\log_0## does not make any sense.
    Second, if your base is positive, your input value may not be negative, ## \forall a,b>0 \log_a (-b) ## does not exist.
    However, I don't think that either of these is the cause of your trouble.

    I would recommend rewriting this using some log rules.

    Edit: incorrect.

    Remove the outer log by putting both sides of the inequality in the exponent.

    Edit: incorrect.

    Then take another look.
     
    Last edited: Dec 9, 2015
  4. Dec 9, 2015 #3

    Mark44

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    Please don't overly decorate your text with various font and BBCode tags if you're not sure of what you're doing. In particular, don't mix BBCode bold tags ( [B] ) with LaTeX -- it screws up both.
     
  5. Dec 9, 2015 #4

    SammyS

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  6. Dec 9, 2015 #5
    So ##(2^{x^2 + 2x + 1} - 1)^y = log_{2x^2 +2x +3 }(x^2 - 2x) - (x^2 + 6x + 10) ##
     
  7. Dec 9, 2015 #6

    SammyS

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    That's not correct.

    ##\displaystyle \frac{log_{\,b}(A)}{log_{\,b}(C)}\neq log_{\,b}(A)-log_{\,b}(C)\ ##

    However, the change of base formula may come in handy.
     
  8. Dec 9, 2015 #7
    ##\frac{log_{2^{x^2+2x+1}-1}(log_{2x^2 + 2x + 3}(x^2 - 2x)}{log_{2^{x^2+2x+1}-1}(x^2 + 6x + 10)} ##
    This was rewritten correctly i think
    ##log_{2^{x^2+2x+1}-1}((log_{2x^2 + 2x + 3}(x^2 - 2x) - (x^2 + 6x + 10)) ##
     
  9. Dec 9, 2015 #8

    RUber

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    edit: incorrect.
     
    Last edited: Dec 9, 2015
  10. Dec 9, 2015 #9

    SammyS

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    Yes. It's worse than I thought. -- that is if you consider one incorrect form to be worse than another.

    What RUber seems to be saying in post #2 is that ##\displaystyle \ \frac{\log_{\,b}(A)}{\log_{\,b}(C)}= \log_{\,b}(A-C)\ ##, which is absolutely wrong.

    Added in Edit:

    What is the case, is that ##\displaystyle \ \log_{\,b}\left(\frac{A}{C}\right)= \log_{\,b}(A)-\log_{\,b}(C)\ ##.

    Also true is, ##\displaystyle \ \frac{\log_{\,b}(A)}{\log_{\,b}(C)}= \log_{\,C}(A)\ ##
     
    Last edited: Dec 9, 2015
  11. Dec 9, 2015 #10

    RUber

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    Wow-- thanks SammyS. I clearly was not paying attention.
    So then using the change of base to remove the quotient, and then exponentiating to remove the logs would be one way to resolve this.
    The other way would be to compare signs of the top and bottom -- but I think that simplifying would help.
     
  12. Dec 9, 2015 #11

    SammyS

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    RUber, you're right. Considerable simplification can be done. Also, as you point out, attention must be paid to several details.

    The "outer' logarithm in numerator has the same base as the logarithm in the denominator, so a change of base is in order. That leaves us with the log of a log to compare with zero.

    As you point out, the base of any of these logarithms must be positive and must not equal 1 . That does eliminate at least one x value and may give OP his value for b, the deleted point.
     
  13. Dec 10, 2015 #12
    So you suggest rewritting to ##\log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0##
     
  14. Dec 10, 2015 #13

    RUber

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    That's right. And solve that like you did for the polynomial in your first post. Pay attention to the inequality. Finally, check for any values that must be excluded from the original form of the problem.
     
  15. Dec 10, 2015 #14
    i went on like this ##\log_{x^2 + 6x + 10} ( \log_{2x^2 + 2x + 3} (x^2 - 2x) ) \ge 0##, ##\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1##, ##2x^2 + 2x + 3 \ge x^2 - 2x##, ##x_1 = -3, x_2 = -1##, also from the change of base formula it is necessary that ##x^2 + 6x + 10 \neq 1## giving same results and that ##2^{x^2 +2x + 1} - 1 \neq 1 ## giving ##x_3=0, x_4=-2##
    what to do with these results?
     
  16. Dec 10, 2015 #15

    SammyS

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    What does ## \displaystyle \ 2^{x^2 +2x + 1} - 1 = 1\ ## imply regarding your initial expression (prior to doing the charge of base)?
     
  17. Dec 10, 2015 #16
    it implies ##\log_{2x^2 + 2x + 3} (x^2 - 2x) = x^2 + 6x + 10 = 1## the base is 1, so it must be 1.
     
  18. Dec 10, 2015 #17

    SammyS

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    But log1 is not defined as a function.
     
  19. Dec 10, 2015 #18

    RUber

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    Was the appropriate range for the inequality ##(x_1, x_2)## or ##( -\infty , x_1)\cup ( x_2, \infty)##?
    Depending on which interval you choose, only one of ##x_3, x_4## would be excluded from the solution set.
    Edit
    It looks like you have the wrong inequality...
    ##\log_{2x^2 + 2x + 3} (x^2 - 2x) \ge 1##, and ##2x^2 + 2x + 3 \ge x^2 - 2x## are not consistent.
    To remove the log, you are doing:
    ##(2x^2 + 2x + 3)^{\log_{2x^2 + 2x + 3} (x^2 - 2x)} \ge (2x^2 + 2x + 3)^1##
    Which should give you:
    ##x^2 - 2x \ge 2x^2 + 2x + 3##.
    This will give you the appropriate interval for the solution once you remove your excluded point.
     
  20. Dec 11, 2015 #19
    you are right, i have made a wrong inequality. so ##x^2 - 2x \ge 2x^2 + 2x + 3## giving me ##x_1 = -1, x_2=-3##. now since i get from the change of base formula that ##x^2 - 2x > 1## and i have what i have written above can i put the x-es i get to see which is excluded? i get by those means the ##-1, -2, -3## fit while ##0## should be excluded. I dont know if you can do it this way
     
  21. Dec 11, 2015 #20
    then it implies that the initial inequality involves undefined functions...what else could it be
     
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