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Math Help - Need help solving another exponential problem.

  1. #1
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    Need help solving another exponential problem.

    So I have the problem:

    16^(d-4) = 3^(3-d)

    My try was this:

    (d-4)log(16) = (3-d)log(3)
    (d-4)log16/log(3) = -(d-3)//Did not want 3-d so I made it d-3 with a negative in front.
    -2.52(d-4) = d-3 //(-2.52 = log(16) / log(3)
    -2.52d + 10.09 = d - 3
    10.09 = -1.52d-3
    13.09 = -1.52d
    d = -8.594

    But the answer in the back of the book says: d = 3.7162

    Where did I go wrong?
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  2. #2
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    You need to ADD \displaystyle 2.52d to both sides...
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  3. #3
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    Quote Originally Posted by thyrgle View Post
    So I have the problem:

    16^(d-4) = 3^(3-d)

    My try was this:

    (d-4)log(16) = (3-d)log(3)
    (d-4)log16/log(3) = -(d-3)//Did not want 3-d so I made it d-3 with a negative in front.
    -2.52(d-4) = d-3 //(-2.52 = log(16) / log(3)
    -2.52d + 10.09 = d - 3
    10.09 = -1.52d-3
    13.09 = -1.52d
    d = -8.594

    But the answer in the back of the book says: d = 3.7162

    Where did I go wrong? ... getting out your calculator too soon and messing with the signs.
    16^{d-4} = 3^{3-d}

    (d-4)\log{16} = (3-d)\log{3}

    d\log{16} - 4\log{16} = 3\log{3} - d\log{3}

    d\log{16} + d\log{3} =  4\log{16} + 3\log{3}<br />

    d(\log{16} + \log{3}) = 4\log{16} + 3\log{3}

    d = \dfrac{4\log{16} + 3\log{3}}{\log{16} + \log{3}}

    now get out your calculator.
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