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.