Urgent: tough equation

quinot · #1 $Old$ November 10th, 2008, 03:37 AM

Hello everybody, I need some help to solve the following equation:

(x^α)*((10-x)^(1-α)) = 10αδ with α between 0 and 1 and δ constant

Thanks a lot!

vincisonfire · #2 $Old$ November 10th, 2008, 06:34 AM

$x^{\alpha} \cdot (10-x)^{1-\alpha} = 10ab$

$x^{\alpha} \cdot (10-x)^{1-\alpha} \cdot (10-x)^{\alpha}= 10ab \cdot (10-x)^{\alpha}$

You know that $\alpha$ is form $\frac{1}{c}$ since $0 < \alpha < 1$

$x^{\alpha} \cdot (10-x)^{1}= 10ab \cdot (10-x)^{\alpha}$

$x \cdot (10-x)^{c}= 10^c a^c b^c \cdot (10-x)$

Know if you would know $\alpha$ you could know c and you could expand and solve the equation analytically if c would be smaller than 5.