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Math Help - Two logarithmetic problems, solve for x

  1. #1
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    Two logarithmetic problems, solve for x

    Attached are the problems.

    For the first problem, the x cancels out each other. So how do I solve for x?

    For the second problem, I don't know how to make the x^-3 into a positive exponent so I can cube root both sides.
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  2. #2
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    Portion 1

    (x+2)log5 = (3+x)log9
    x+2/3+x = log9/log5
    x+2/3+x = 1.37 ( cross multiply )
    x+2 = 4.11 + 1.37x
    -0.37x = 2.11
    x = -5.7

    2nd portion

    x^-3 = 125
    1 / x^3 = 125
    x^3 = 1/125
    x = Cube root(1/125)
    x = 1/5
    Last edited by tester85; December 18th 2008 at 09:40 AM.
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  3. #3
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    How did you do the second portion? Like what do I do when there is a negative exponent and I have to take the root of both sides?
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  4. #4
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    x^-3 = 125
    1 / x^3 = 125
    x^3 = 1/125
    x = Cube root(1/125)
    x = 1/5
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  5. #5
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    Quote Originally Posted by mwok View Post
    For the second problem, I don't know how to make the x^-3 into a positive exponent so I can cube root both sides.

    Well, this is how you do the second part

    x^{-3} = 125

    Remeber that x^{-3}= \frac{1}{x^{3}}

    \frac{1}{x^{3}} = 125

    Now cross multiply bth sides

    x^{3} = \frac{1}{125}

    Take the cube root on both sides and you get

    x = \frac{1}{5}
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  6. #6
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    Ahh, thanks!
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