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Math Help - Algebra with huge exponents

  1. #1
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    Algebra with huge exponents

    4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)

    Solve for X
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  2. #2
    Super Member wingless's Avatar
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    4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^x)
    4^1993(4^3 - 4^2 - 4^1 + 4^0) = 45(2^x)
    2^3986*45 = 45*(2^x)
    x = 3986
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  3. #3
    Bar0n janvdl's Avatar
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    Quote Originally Posted by stones44 View Post
    4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)

    Solve for X
    4^{1996} = 4^3 (4^{1993})

    4^{1995} = 4^2 (4^{1993})

    4^{1994} = 4 (4^{1993})

    Set y = (4^{1993})

    Then we have:

    64y - 16y - 4y + y = 45(2^x)

    45y = 45(2^x)

    y = 2^x

    4^{1993} = 2^x

    2^{2 \times 1993} = 2^x

     x = 3986
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  4. #4
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    Quote Originally Posted by stones44 View Post
    4^1996 - 4^1995 - 4^1994 + 4^1993 = 45(2^X)

    Solve for X
    4^{1996}-4^{1995}-4^{1994}+4^{1993} = 4^{1993}(4^3  - 4^2 - 4 + 1) = 45\cdot 2^x
    Thus,
    45 \cdot 4^{1993} = 45 \cdot 2^x \implies 4^{1993} = 2^x.
    Now convert both sides to base 2 and finish the problem.
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  5. #5
    Bar0n janvdl's Avatar
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    Hehehe, 3 answers to one problem. At least all of them are right.
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  6. #6
    GAMMA Mathematics
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    Quote Originally Posted by janvdl View Post
    Hehehe, 3 answers to one problem. At least all of them are right.
    That was great guys -- working together in harmony!
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  7. #7
    Bar0n janvdl's Avatar
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    Quote Originally Posted by colby2152 View Post
    That was great guys -- working together in harmony!
    I would say it was simply all of us posting at the same time...
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