# Solutions of $\frac{1- x^{\alpha}}{1- x^{\beta}} = \gamma$

Would it be possible to give pieces of information about $x \in \left]0,1\right[$ such that:

$$\frac{1- x^{\alpha}}{1- x^{\beta}} = \gamma$$

where $\alpha > 0$, $\beta > 0$ and $\gamma > 0$.

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Assume without loss of generality that $\alpha\lt\beta$. Then the function $u:x\mapsto(1-x^\alpha)/(1-x^\beta)$ is decreasing on the interval $[0,1]$ from $u(0)=1$ to $u(1)=\alpha/\beta$. Hence the equation $u(x)=\gamma$ has exactly one solution $x$ in $[0,1]$ if $\alpha/\beta\leqslant\gamma\leqslant1$, and no solution otherwise.
To determine the variation of $u$, note that $u'(x)$ has the sign of $$v(x)=\beta x^{\beta-\alpha}-(\beta-\alpha) x^\beta-\alpha,$$ and that $v'(x)=\beta(\beta-\alpha)x^{\beta-\alpha-1}(1-x^\alpha)$ hence $v'(x)\gt0$ for every $x$ in $(0,1)$. Since $v(1)=0$, this proves that $v(x)\lt0$ for every $x$ in $(0,1)$, hence $u$ is decreasing.
Thank you! A quick note: you meant $\alpha/\beta\leq\gamma\leq1$ at the end of the first paragraph. –  Wok Sep 26 '12 at 15:51
What about using $x=u^{1/\beta}$ to get $$f(u)=\frac{1-u^{\alpha/\beta}}{1-u}=\gamma\;.$$ $f(u)$ ranges between $1$ and the maximal value, given by $\lim_{u\to 1}f(u)=\frac\alpha\beta$, which implies $\frac\alpha\beta \ge\gamma$.
Thank you! A quick note: you implicitly assumed $\alpha\geq\beta$. –  Wok Sep 26 '12 at 15:52