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Math Help - Make x the subject

  1. #1
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    Make x the subject

    Make x the subject of the formula: v=\frac{1}{k}\sqrt{[1+(\frac{1}{x}+1)^2}/Lg In case I made a mistake in the coding, its v=1/k sqrt(1+(1/x +1)^2/Lg)
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  2. #2
    Senior Member tukeywilliams's Avatar
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     \text{Lg} as in the logarithm with base 2? The formula doesn't make sense (i.e. the denominator).
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  3. #3
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    No Lg are just two different letters. The second part of the question shows this when they ask you to put in values for each. k = 1.26, L = 48.55, g = 32.21, v = 0.0483 and solve for x.
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  4. #4
    Senior Member tukeywilliams's Avatar
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    So we have  v = \frac{\sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}}{kLg}

    Then  vkLg = \sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}

    Or  v^{2}k^{2}L^{2}g^{2}-1 = \left(\frac{1}{x} +1 \right)^2

    So   \left(\frac{1}{x} +1 \right)^2 = \frac{1}{x^2} + \frac{2}{x} +1 and  v^{2}k^{2}L^{2}g^{2}-2 = \frac{1}{x^2} + \frac{2}{x}
    Last edited by tukeywilliams; August 3rd 2007 at 01:01 AM.
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  5. #5
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by tukeywilliams View Post
    So we have  v = \frac{\sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}}{kLg}

    Then  vkLg = \sqrt{\left[1 + \left(\frac{1}{x}+1 \right)^{2} \right]}

    Or  v^{2}k^{2}L^{2}g^{2}-1 = \left(\frac{1}{x} +1 \right)^2
    Actually it's simpler at this point to use:
    \frac{1}{x} + 1 = \pm \sqrt{v^{2}k^{2}L^{2}g^{2}-1}

    and solve for x from there.

    -Dan
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