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Proof/Show using index laws

  1. Mar 1, 2013 #1
    1. The problem statement, all variables and given/known data
    Show that the following can be expressed as a quadratic equation

    [tex]2^{x}-1 = \frac{2}{2^{x}}[/tex]


    2. Relevant equations
    Show that the quadratic has only one real solution at [itex]x = 1[/itex]



    3. The attempt at a solution
    Well, my biggest problem is the main part of the task, converting the equation into a quadratic. I cannot find a way to get the same base for all parts of the equation so I can equate them. I've tried doing things like:
    -Taking the denominator to the top on the RHS and making [itex]1 = 2^{0}[/itex]
    - Squaring both sides of the equation, although I can get a quadratic out of this method, technically I haven't proved anything, just made the equation look a little fancier.
    - taking the -1 to the other side and adding it in the form of [itex]\frac{2^{x}}{2^{x}}[/itex]

    And others too, but those were the ones which actually led me somewhere (but that somewhere was a dead end).

    Can anyone give me any hints on how to approach a question like this?

    ps. I've also tried not to assume x = 1 while doing my conversion to a quadratic.
     
    Last edited by a moderator: Mar 1, 2013
  2. jcsd
  3. Mar 1, 2013 #2

    rock.freak667

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    Homework Helper

    The 2^x in the denominator doesn't look too helpful in that position, so make it look nicer and multiply throughout by 2^x :wink:
     
  4. Mar 1, 2013 #3

    HallsofIvy

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    Much the same thing: replace each "[itex]2^x[/itex]" with "y" and then multiply both sides by y.
     
  5. Mar 2, 2013 #4
    I tried the multiply y to both sides, but I end up getting something like this:

    [itex]\frac{2^{2x}-2^{x}}{2^{x}}=\frac{2^{x+1}}{2^{2x}}[/itex]

    Which is where I get stuck again. Although I get a quadratic when I sub in the y, when I revert it back to [itex]2^{x}[/itex], the function becomes a plain old exponential function.

    If I split the LHS at the "-" I'll still get the pesky -1 and ending up cancelling out one of the x, which put me at a dead end :cry:

    How do I proceed?
     
  6. Mar 2, 2013 #5
    Hey you did the multiplication wrong! ##2^{x+1}##??

    It should be: ##2^{2x}-2^x=2##. Now a simple substitution of 2x=y will give you a quadratic equation..
     
  7. Mar 2, 2013 #6
    I multiplied both sides by
    [itex]\frac{2^{x}}{2^{x}}[/itex] that's how I got that result

    Yeah but is that really a quadratic equation? or an exponential function in disguise. Because if I have my quadratic in terms of y, as soon as I sub in [itex]2^{x}[/itex] it turns into an exponential function. I checked this by graphing it, and it does not pass through x = 1

    I graphed:
    [itex]2^{2x}-2^{x}+2 = y[/itex]
     
  8. Mar 2, 2013 #7
    No! You don't have to graph that.
    Solve: ##y^2-y-2=0##, where y=2x..

    Then you get 2 values for ##2^x##, using which you can prove the next part of the question.
     
  9. Mar 2, 2013 #8
    Ohhh... I didn't even think of that :biggrin: I think I've got this now.

    Thank you, you've made my day :rofl:
     
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