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Find the range of solutions

  1. Dec 6, 2015 #1

    diredragon

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    1. The problem statement, all variables and given/known data
    ##4^{x} - 7*2^{(x-3)/2} = 2^{-x} ##
    Set of real solutions of this equation is found in which following range:
    a) (-9, -2)
    b) (0, 3)
    c) (-3, 0]
    d) (3, 7]
    2. Relevant equations
    3. The attempt at a solution

    I simplified to
    ##2^{3x} - 7*2^{(3x - 3)/2} = 1##
    ##2^{3x} - \frac{7*2^{1/2}}{4}2^{3x/2} - 1 = 0 ##
     
    Last edited by a moderator: Dec 6, 2015
  2. jcsd
  3. Dec 6, 2015 #2

    haruspex

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    There is something that every occurrence of x has in common. This suggests a change of variable.
     
  4. Dec 6, 2015 #3

    diredragon

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    This came to mind
    ##z^2 = 2^{3x} ##
    ##z^2 - \frac{7*2^{1/2}}{4}z - 1 = 0 ##
    solutions of this equation i named q and t
    ##q = 2*2^{1/2} ##
    ##t = \frac{-1}{4}2^{1/2} ##
    I then get two values of ##x ##, ##1 ## and ##-1 ##
     
    Last edited: Dec 6, 2015
  5. Dec 6, 2015 #4

    haruspex

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    Have you checked both of those satisfy the original equation?
     
  6. Dec 6, 2015 #5

    diredragon

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    I get that neither satisfys the equation. What is the mistake?
     
  7. Dec 6, 2015 #6

    haruspex

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    One does. The other came in because the use of z2 created an ambiguity.
     
  8. Dec 6, 2015 #7

    diredragon

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    Oh I didn't see. ##1 ## fits. But I don't see how I can get the range which is asked. 1 is found in only one given answer so the solution i guess can only be (0, 3]
     
    Last edited: Dec 6, 2015
  9. Dec 6, 2015 #8

    haruspex

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    Looks right.
     
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