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Math Help - solving for unknowns

  1. #1
    Junior Member
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    solving for unknowns

    1.) If,
    {4x^2 - sqrt[3]{x^2} + 3}/{sqrt[6]{x^5}} = 4x^a - x^b + 3x^c}

    then a= ______ , b= _______ , and c= _______ .

    2.) If V=4r^2h and r/(H-h)= R/H, write V as a function of R
    V= _______________


    3) Solve for
    S= (a^n) -1 /i

    a= __________

    4.) Solve for n:
    P= R[(1-x^-n)/i]
    n=___________
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  2. #2
    Super Member Matt Westwood's Avatar
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    I'll give you a headstart on the first one ...
    <br />
\frac {4x^2 - \sqrt[3]{x^2} + 3}{\sqrt[6]{x^5}} = 4x^a - x^b + 3x^c<br />

    Multiply through by that \sqrt[6]{x^5} for a start, and use fractional indices to make it easier to read:

    <br />
4x^2 - x^{2/3} + 3 = x^{5/6}\left({4x^a - x^b + 3x^c}\right)<br />

    Then equate the terms with coefficients 4, -1 and 3 and remember that when multiplying two terms you add their indices.
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  3. #3
    MHF Contributor
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    1.) If,
    {4x^2 - sqrt[3]{x^2} + 3}/{sqrt[6]{x^5}} = 4x^a - x^b + 3x^c}
    then a= ______ , b= _______ , and c= _______ .


    Does sqrt[3]{x^2] mean cuberoot of x^2? And sqrt[6](x^5) the 6th root of x^5?
    If yes, then,
    Rewriting your original equation,
    [4x^2 -x^(2/3) +3] / [x^(5/6] = 4x^a -x^b +3x^c
    4x^(2 -5/6) -x^(2/3 -5/6) +3x^(-5/6) = 4x^a -x^b +3x^c
    4x^(7/6) -x^(-1/6) +3x^(-5/6) = 4x^a -x^b +3x^c

    Therefore, a = 7/6; b = -1/6; and c = -5/6 ------------answer.

    ---------------------------------------------------------------
    2.) If V=4r^2h and r/(H-h)= R/H, write V as a function of R
    V= _______________


    The idea here is to put V and R into one equation only.

    We see that h is common in the two given equations, so we work on that:
    From 1st equation, h = V/(4r^2).
    Substitute that into the 2nd equation,
    r /[H -V/(4r^2)] = R/H.
    Cross multiply,
    rH = R[H -V/(4r^2)]
    rH = RH -VR/(4r^2)
    Isolate V,
    VR/(4r^2) = RH -rH
    V = (RH -rH) /[R/(4r^2)]
    V = (4r^2)(RH -rH) /R
    V = (4Hr^2)(R-r) /R -------------answer.

    Or, V = (4Hr^2)(1 -r/R)
    ------------------------------------------------

    3) Solve for
    S= (a^n) -1 /i

    a= __________


    The idea here is to just isolate a.

    iS = a^n -1
    a^n = iS +1
    Take the log of both sides,
    n*log(a) = log(iS +1)
    n = log(is +1) / log(a) -------------answer.

    ------------------------------------------------------------------------
    4.) Solve for n:
    P= R[(1-x^-n)/i]
    n=___________


    Just isolate n:

    P/R = (1 -x^(-n) / i
    iP/R = 1 -x^(-n)
    x^(-n) = 1 -iP/R
    Take the log of both sides,
    (-n)log(x) = log(1 -iP/R)
    -n = log(1 -iP/R) / log(x)
    n = -log(1 -iP/R) / log(x) -------------answer.

    Can also be:
    n = -log[(R -iP)/R] / log(x)
    n = [-log(R -iP) +log(R)] / log(x)
    n = [log(R) -log(R -iP)] / log(x) ------answer also.
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