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Find the sets of real solutions

  1. Jan 9, 2016 #1
    [b[1. The problem statement, all variables and given/known data[/b]

    ##|4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x| \ge 8*6^x(8^{x-1}+6^x)##

    The sets containing the real solutions for some numbers ##a, b, c, d,## such that ##-\infty < a < b < c < d < +\infty## is of the form ##(-\infty, a] \cup [b, c] \cup [d, +\infty)##. Prove it by obtaining the solution set.

    2. Relevant equations

    3. The attempt at a solution

    So the problem is asking me to solve the inequality by finding some numbers ##a, b, c, d## and find the form of the real solution set that should be in the form given above.

    Here is my current work:

    1) ##4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x \geq 8*6^x(8^{x-1}+6^x) ##

    2) ##4^{3x}-2^{4x+2}*3^{x+1}+20*12^x*3^x \leq -8*6^x(8^{x-1}+6^x) ##

    1)

    ##2^{6x}-2^{4x+2}*3^{x+1}+20*2^{2x}*3^{2x} \geq 8*3^x*2^x(2^{3x-3}+2^x*3^x) ##

    ##2^{6x}-12*2^{4x}*3^{x}+20*2^{2x}*3^{2x} \geq 8*3^{2x}*2^{2x}+8*3^{x}*2^{4x-3} ##

    ##2^{6x}-12*2^{4x}*3^{x}+12*2^{2x}*3^{2x} \geq 3^{x}*2^{4x} ##

    ##2^{6x}-13*2^{4x}*3^{x}+12*2^{2x}*3^{2x} \geq 0 ##

    ##2^{4x}-13*2^{2x}*3^{x}+12*3^{2x} \geq 0 ##

    2)

    ##2^{4x}-11*2^{2x}*3^{x}+28*3^{2x} \leq 0 ##

    not sure now how to continue, am i on the right track? (latex: messed up something again, cant find what.)
     
    Last edited by a moderator: Jan 10, 2016
  2. jcsd
  3. Jan 9, 2016 #2

    haruspex

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    See if you can factorise that.
     
  4. Jan 9, 2016 #3
    I can factor out ##3^x## out of the last two. ##2^{4x} - 3^x(13*2^{2x} + 3^x) \geq 0##
     
  5. Jan 9, 2016 #4

    haruspex

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    Right, and do you notice a common factor of the first two? Does that suggest a way of factorising the whole?
    It might help if you substitute a for 2x and b for 3x.
     
  6. Jan 10, 2016 #5
    ##a^4 - b(13*a^2 + b) \geq 0##
    Sorry, but i dont see a way to fact out the whole thing... what do ##a^4## and ##b## have in common. This kind a looks like a quadratic if i set ##g= a^2## and if ##b=constant##. Other than that i dont know
     
  7. Jan 10, 2016 #6

    haruspex

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    Sorry, I made two mistakes. I didn't notice some errors in your post #3, and I meant to say write a for 4x. Going back to the original form but using the a, b substitution, you have a2-13ab+12b2.
    If you can't spot the factors just apply the quadratic formula.
     
  8. Jan 10, 2016 #7
    How can i apply the formula when there are two square terms in here?
    I tried this:
    ##x_1=\frac{13ab + \sqrt{169a^2b^2 - 48b^2}}{2}## is this right?
     
  9. Jan 10, 2016 #8

    haruspex

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    No. You have been inconsistent regarding the role of a. In some terms you have treated it as the x in a standard quadratic, in other terms as a coefficient. Try it again.
     
  10. Jan 10, 2016 #9
    oops, i see. so ##x_1=\frac{13b + \sqrt{169b^2 - 48b^2}}{2}##, ##x_1=\frac{13b - \sqrt{169b^2 - 48b^2}}{2}##, ##x_1=12b##, ##x_2=b##
    ##(a - 12b)(a-b) \geq 0##
    ##(4^x - 12*3^x)(4^x-3^x) \geq 0##
    i now want to solve for ##=0##
    ##(4^x - 12*3^x)(4^x-3^x) = 0##
    ##(4^x - 12*3^x)=0## or ##(4^x-3^x) = 0## from this i only see ##x_1=0## as the solution to ##(4^x-3^x) = 0## and thats because its obvious, the first one however i cant find x.
     
  11. Jan 10, 2016 #10

    haruspex

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    You can make the 12 go away by merging it into the power terms. Then take logs.
     
  12. Jan 10, 2016 #11

    Mark44

    Staff: Mentor

    @diredragon, please don't try to mix BBCode bold tags with the LaTeX markup. They don't play well together.

    I fixed your first post here, and this was the second time that I fixed the mess that you made. Please try to be more careful in the future.
     
  13. Jan 12, 2016 #12
    Do you mean i can make it to ##4^x - 4*3^{x+1} = 0##, ##4(4^{x-1} - 3{x+1}) = 0##, how now to take logs? Log the both sides? Im not sure what to do.
     
  14. Jan 12, 2016 #13

    haruspex

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    I assume you meant ##4(4^{x-1} - 3^{x+1}) = 0##.
    Since it is zero, you can drop the factor 4. Then move one term to the other side and take logs.
     
  15. Jan 12, 2016 #14
    ##(x-1)log_e(4) =(x+1)log_e(3)##
    ##xlog_e(4) - xlog_e(3)=log_e(3) + log_e(4)##
    ##x= \frac{log_e(3) + log_e(4)}{log_e(4) - log_e(3)}##
     
  16. Jan 12, 2016 #15

    haruspex

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    Bingo.
     
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