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Math Help - Progressions

  1. #1
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    Progressions

    If 1, log x (base y), log y (base z), -15 log z (base x)
    are in AP, then
    a) z^3 = x
    b) x = y^-1
    c) z^-3 = y
    d) all of these

    Regards
    Sreedhar
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  2. #2
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    Let the common difference be d. Then the progression is 1, d+1, 2d+1, and 3d+1, so
    x=y^{d+1}
    y=z^{2d+1}
    z=x^{-(3d+1)/15}\text{ or }x=z^{-15/(3d+1)}
    Combining the first two gives x=z^{(2d+1)(d+1)}, and equating exponents gives (3d+1)(2d+1)(d+1)=-15, with the only solution d=-2.

    - Hollywood
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  3. #3
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    Hello, yshridhar!

    Lovely work, Hollywood!


    \text{If }\;1,\;\log_y(x),\;\log_z(y),\;-15\log_x(z)\,\text{ are in AP, then:}

    . . (a)\;z^3 = x \qquad (b)\;x = y^{-1} \qquad (c)\;z^{-3} = y \qquad (d)\;\text{all of these}

    Hollywood had: . \begin{Bmatrix}[1] & x &=& y^{d+1} \\ [2] & y &=& z^{2d+1} \\ [3] &z &=& x^{-\frac{3x+1}{15}}\end{Bmatrix}\quad\text{ and }\:d =- 2


    Then we have:

    . . [1]\;\;x \:=\:y^{-1} \;\;{\color{blue}(b)}

    . . [2]\;\;y \:=\:z^{-3} \quad\Rightarrow\quad z^{-3} \:=\:y \;\;{\color{blue}(c)}

    . . [3]\;z \:=\:x^{\frac{1}{3}} \quad\Rightarrow\quad z^3\:=\:x\;\;{\color{blue}(a)}


    Answer: . {\color{blue}(d)\;\text{ all of these}}

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  4. #4
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    Thanks alot Hollywood for such a simple solution. I am banging with logarithms formulae for this sum.
    Regards
    Sridhar
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