Algebra question

**mabruka** · April 20th 2010, 01:01 PM

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

**Deadstar** · April 20th 2010, 02:13 PM

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

$(3-2)\log(\tfrac{3}{2}) = 0.4054651081$

**Soroban** · April 20th 2010, 02:38 PM

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

**mabruka** · April 20th 2010, 03:13 PM

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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