Thread: Urgent: tough equation

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!

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

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

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

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

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

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