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Find the solution set of 2^(2x-2)-2*2^(x-1)=8

  1. May 11, 2015 #1
    1. The problem statement, all variables and given/known data
    Find the solution set of 2^(2x-2)-2*2^(x-1)=8

    2. Relevant equations


    3. The attempt at a solution
    2^(2x-2)-2*2^(x-1)=8
    2^(2x-2)-2^(x-1+1)=8
    2^(2x-2)-2^(x)=8
    2^(2x-2)-2^(x)-8=0
    2^(2(x-1))-2^(x)-8=0
    I cannot solve the equation,

    I just need direction with this step and will attempt the rest on my own, thank you.
    Jaco
     
    Last edited: May 11, 2015
  2. jcsd
  3. May 11, 2015 #2

    BvU

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    Re grasping: you can do this: 2*2(x-1) = 2x.
    And you can do something with 22x too ! Once you see that, it's easy ...
     
  4. May 11, 2015 #3
    Thank you BvU

    2^(2x-2)-2*2^(x-1)=8
    2^(2x-2)-2^(x-1+1)=8
    2^(2x-2)-2^(x)=8
    2^(2x-2)-2^(x)-8=0
    2^(2(x-1))-2^(x)-8=0
    4^(x-1)-2^x=8

    or
    2^(2x-2)-2*2^(x-1)=8
    2^(2(x-1))-2*2^(x-1)=8
    4^(x-1)-2*2^(x-1)=8
    4^(x-1)-(2*2^(x-1))=8
    4^(x-1)-2^(x-1+1)=8
    2^(2(x-1))-2^x=8
    (2x-1)2-2x=8
    ax2+bx+c=0
    (2x-1)2-2x-8=0
     
    Last edited: May 11, 2015
  5. May 11, 2015 #4
    Hi Jaco,
    You should use either the sub-script and super-script buttons from the post template or ##LATEX##, it will make your posts a lot more readable.
    The question is basically asking which two powers of 2 have the difference 8. If you just want an answer look at the differences of the first few powers of 2.
    To prove it algebraically, you could either factor the expression ##2^{2x-2}-2^x## or continuing on what BvU said, modify the expression ##4^{x-1}## a little more and it should become familiar territory.
     
  6. May 11, 2015 #5

    HallsofIvy

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    The first thing I would do is let [itex]y= 2^{x- 1}[/itex].

    [mentor note: content abridged]

    ...... see how you go with that substitution.
     
    Last edited by a moderator: May 11, 2015
  7. May 11, 2015 #6

    BvU

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    :smile: Yes 22x = (22)x. But it's also 2x2 = ... Aha !
     
  8. May 11, 2015 #7
    Hi Hallsoflvy,
    how do you get the 2y? b=-2x not 2x-1
    or should I be stepping back to:
    2(x-1)2-2(2x-1)=8
    ax2+bx+c=0
    y2-2y-8=0
    y2-2y-8=0
     
    Last edited: May 11, 2015
  9. May 11, 2015 #8
    HallsofIvy took ##b=2^{x-1}## BvU and I took ##b=2^{x}##.
     
  10. May 11, 2015 #9
    I also used b=2x
    But I will attempt this way.
     
  11. May 11, 2015 #10
    Hi Certainly,
    where did you get 3?
    I know you see it must be 3, but how is it shown?
    Should I not have proved this?
     
  12. May 11, 2015 #11
    In order for ##2^{x-2}-1>0## the power of 2 must be at-least 1. Therefore ##x-2=1## or ##x=3##.
     
  13. May 11, 2015 #12
    22x-2-2*2x-1=8
    22x-2-2x-1+1=8
    22x-2-2x=8
    2x(2x-2-1)=8
    for 2x(2x-2-1)>0 the power of 2<=1
    x=3
    {3}
    Is that how I would end my calculation?
     
    Last edited: May 11, 2015
  14. May 11, 2015 #13

    SammyS

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    2(x-1)2 is equivalent to (2(x-1))2 . You did get this right, but the extra parentheses makes clear what is intended.

    However, if y = 2x-1 , then 2(2x-1) = 2(y), not 4(y) .
     
  15. May 11, 2015 #14
    Thank you Sammy,
     
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