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Math Help - Random Equation Help?

  1. #1
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    Dec 2008
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    9

    Random Equation Help?

    Obviously, a calculator cannot be used.

    1)
    Code:
    3x^2 + 12x - 18
       ------------------- = 0
              x + 7
    2) 9^x + (4)3^+x - 3 = 0

    3) 5^x = 4^x+5

    4) 2log7(x+6) - log7(x+5) = 2 (the 7's are the log base)

    5) log8(e^x+4) = 2 (8is the log base)


    Rep and thanks to anyone who will solve these!
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  2. #2
    Member
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    Dec 2008
    From
    Auckland, New Zealand
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    189
    (1)

    \frac{3x^2 + 12x - 18}{x + 7} = 0

    \frac{3(x^2 + 4x - 6)}{x + 7} = 0

    To solve this, solve (x^2 + 4x - 6) = 0. This gives the only two solutions (use the quadratic formula)

    (2)Not sure it isn't written clear enough

    (3) Not sure on this one sorry...

    (4)

    2\log_{7} (x + 6) - \log_{7}{(x + 5)} = 2

     \log_{7} [(x + 6)^2] - \log_{7} (x + 5) = 2

    \log_{7} \left(\frac{(x + 6)^2}{x + 5}\right) = 2

    \frac{(x + 6)^2}{x + 5} = 7^2 = 49

    (x + 6)^2 = 49(x + 5)

    Now this is a quadratic equation, solve for x. Remember to check the solutions so that x + 6 > 0 and x + 5 > 0 because you cannot have logs of negative numbers.

    (5)

    \log_{8} (e^x + 4) = 2

    e^x + 4 = 8^2 = 64

    e^x = 60

    x = \log_{e} 60
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  3. #3
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    Dec 2008
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    Thanks for the help!

    One small problem, due to me not knowing how to make the problems look neat..


    number 5 is

    log8(e^(x+4))=2

    the e is raised to the x+4..



    and on number 2

    9^x + (4)3^+x - 3 = 0

    it is 9 raised to the x, plus 4 times 3raised to the +x, -3 = 0, if that helps...

    thanks a lot again!
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  4. #4
    Member
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    Dec 2008
    From
    Auckland, New Zealand
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    189
    I'm not too sure on the equations with different numbers raised to power x.

    But number (5):

    \log_{8} (e^{x + 4}) = 2

    e^{x + 4} = 8^2 = 64

    x + 4 = \log_{e} 64 = \ln 64
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