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Math Help - Values of 2^x

  1. #1
    Junior Member Dragon's Avatar
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    Values of 2^x

    For what values of 2^x does

    2^x+2^x+1=4^x+4^x+1
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  2. #2
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    Quote Originally Posted by Dragon View Post
    For what values of 2^x does

    2^x+2^x+1=4^x+4^x+1
    Let y=2^x

    Thus,

    y+y+1=y^2+y^2+1

    You shall solve.
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  3. #3
    Member Rimas's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    Let y=2^x

    Thus,

    y+y+1=y^2+y^2+1

    You shall solve.

    i keep on getting a negative square root??
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by Dragon View Post
    For what values of 2^x does

    2^x+2^x+1=4^x+4^x+1
    Using TPH's method:

    Let y = 2^x

    Then
    y + y + 1 = y^2 + y^2 + 1

    2y = 2y^2 <-- Cancel the common factor of 2

    y^2 - y = 0

    y(y - 1) = 0

    So y = 0 or y = 1.

    Thus 2^x = 0 <-- Impossible!

    or

    2^x = 1 <-- x = 0

    So the only solution is x = 0.

    -Dan
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  5. #5
    Member Rimas's Avatar
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    Thanks alot
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  6. #6
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    Hello, Dragon!

    Some parentheses and spaces would have helped.
    I'm guessing what the problem is supposed to be . . .


    For what values of 2^x does: .2^x + 2^(x+1) .= .4^x + 4^(x+1)

    If I've interpreted the problem correctly, we get an entirely different answer.


    We have: .2^x + 22^x .= .(2)^x + (2)^(x+1}

    . . 32^x .= .2^(2x) + 2^(2x + 2)

    . . 32^x .= .2^(2x) + 22^(2x)

    . . 32^x .= .2^{2x} + 42^(2x)

    . . 32^x .= .52^(2x)

    Divide by 2^x: .3 .= .52^x . **

    And we have: .2^x .= .3/5

    ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~ ~

    ** . We can divide by 2^x because 2^x ≠ 0.

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