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Math Help - (8^x)+(8^(-x))

  1. #1
    Junior Member Singular's Avatar
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    (8^x)+(8^(-x))

    if (4^x)+(4^(-x))=7

    (8^x)+(8^(-x))= ????

    thanks
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  2. #2
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    Hello, Singular!

    This is a most unplesant problem . . . especially without LaTeX.


    If .4^x + 4^{-x} .= .7, then find: .8^x + 8^{-x}
    The given equation is: .2^{2x} + 2^{-2x} .= .7


    Cube both sides: .[2^{2x} + 2^{-2x}] ,= ,7

    . . 2^{6x} + 3[2^{4x}][2^{-2x}] + 3[2^{2x}][2^{-4x}] + 2^{-6x} .= .343

    . . (2)^{2x} + 3[2^{2x}] + 3[2^{-2x}] + (2)^{-2x} .= .343

    . . 8^{2x} + 3[4^x + 4^{-x}] + 8^{2x} .= .343
    . . . . . . . . . . .\__________/
    . . . . . . . . . . . . .
    This is 7

    So we have: .8^{2x} + 21 + 8^{-2x} .= .343 . . 8^{2x} + 8^{-2x} .= .322

    . . . . . . . . . (8^x) + (8^{-x}) .= .322 . [1]

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

    We will take a slight detour now . . .

    Consider the square of: 8^x + 8^{-x} .= .k

    We have: .[8^x + 8^{-x}] .= .(8^x) + 2(8^x)(8^{-x}) + (8^{-x}) .= .k

    . . . . . . . . (8^x) + 2 + (8^{x}) .= .k . . (8^x) + (8^{-x}) .= .k - 2


    Substitute into [1]: .k - 2 .= .322 . . k = 324 . . k = 18


    Therefore: . 8^x + 8^{-x} .= .18



    Is there a better way? . . . I hope so!

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  3. #3
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    Quote Originally Posted by Soroban View Post
    Hello, Singular!

    This is a most unplesant problem . . . especially without LaTeX.


    2^{2x} + 2^{-2x}=7
    Why do you say that?
    The problem does not say what this is.
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  4. #4
    Forum Admin topsquark's Avatar
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    If
    2^{2x} + 2^{-2x}
    is not equal to
    4^x + 4^{-x}
    then what should it be?

    -Dan
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  5. #5
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    Code:
    8^{x}+8^{-x}=2^{3x}+2^{-3x}
                =[2^{x}+2^{-x}][2^{2x}-1+2^{-2x}]
                =[2^{x}+2^{-x}][4^{x}-1+4^{-x}]
                =[2^{x}+2^{-x}][6]
    
    Now find the value of  
    
    [2^{x}+2^{-x}]^{2}=[2^{2x}+2+2^{-2x}]
                      =9
    
    Thus (3)(6)=18
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