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Math Help - Algebra question

  1. #1
    Member mabruka's Avatar
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    Algebra question

    I have problems figuring out what was done here:




    \frac{p^3 -q^3}{3(1-pq)}\log\frac{p^3}{q^3}   =    (p-q) \log\frac{p}{q}



    How does the left side follow?

    I can see the lograithm really doesnt play an important role here, so the actual question is how


    \frac{p^3 -q^3}{(1-pq)}=p-q

    tahnk you
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  2. #2
    Super Member Deadstar's Avatar
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    Were there any conditions on p and q? Cos that relationship doesn't hold...

    Simple example. Let p=3, q=2.

    \frac{3^3 - 2^3}{3(1-6)}\log(\tfrac{3^3}{2^3}) = -1.540767411

    But

    (3-2)\log(\tfrac{3}{2}) = 0.4054651081
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  3. #3
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    Hello, mabruka!

    I agree with deadstar . . . something is missing.

    Perhaps these are probabilities? . {\color{blue}p + q \:=\:1}

    \frac{p^3 -q^3}{3(1-pq)}\cdot\log\left(\frac{p^3}{q^3}\right) \;=\; (p-q)\cdot \log\left(\frac{p}{q}\right)

    How does the right side follow?

    The left side is: . \frac{p^3-q^3}{3(1-pq)}\cdot\log\left(\frac{p}{q}\right)^3 \;=\;\frac{p^3-q^3}{{\color{red}\rlap{/}}3(1-pq)}\cdot {\color{red}\rlap{/}}3\cdot\log\left(\frac{p}{q}\right)

    . . . . . . . . . . =\;\frac{p^3-q^3}{1-pq}\cdot\log\left(\frac{p}{q}\right) \;=\;\frac{(p-q)(p^2+pq + q^2)}{1-pq}\cdot\log\left(\frac{p}{q}\right) .[1]



    If p + q \:=\:1, then: . (p+q)^2 \:=\:1^2 \quad\Rightarrow\quad p^2 + 2pq + q^2 \:=\:1 \quad\Rightarrow\quad p^2+pq + q^2 \:=\:1-pq


    Substitute into [1]: . \frac{(p-q)({\color{red}\rlap{//////}}1-pq)}{{\color{red}\rlap{//////}}1-pq}\cdot\log\left(\frac{p}{q}\right) \;=\;(p-q)\cdot\log\left(\frac{p}{q}\right)

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  4. #4
    Member mabruka's Avatar
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    YES!! I totally forgot about that little detail when i was verifying it.
    They are complementary probabilities, so their sum is 1!




    Thank you both !
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