I am struggling with what I feel is not a particularly difficult problem. I want to show that
$$ f(x) = \frac{b}{A^{b}} x^{b} - \frac{a}{A^{a}} x^{a} - 1 $$
has exactly one root for $b>a>0$, $A>0$ and $x \geq 0$ for all real $x$. If $b$ and $a$ are integers it is easy to show since the function becomes a polynomial and I can apply Descartes rule of signs. Moreover, if $a$ and $b$ are rational, I can simply replace $z^{p} = x$ where $p$ is the sum of the denominators of $a$ and $b$ and once again apply Descartes rule of signs to the polynomial. However, I have not found a way to go about it for all nonnegative real $x$.
Proving the existence of at least one solution is fine, I simply apply the intermediate value theorem, but I cannot find a way to show that this solution is unique. Would appreciate any guidance that I can get!
\begin{align} f'(x)&=\frac{b^2}{A^b}x^{b-1}-\frac{a^2}{A^a}x^{a-1}\\ &=b^2A^{-a}x^{a-1}\left(\left(\frac{x}{A}\right)^{b-a}-\frac{a^2}{b^2}\right)\\ &\cases{<0&for $\ x<A\,\sqrt[b-a]{\frac{a^2}{b^2}}$\\ =0&for $\ x=A\,\sqrt[b-a]{\frac{a^2}{b^2}}$\\ >0&for $\ x>A\,\sqrt[b-a]{\frac{a^2}{b^2}}$\\} \end{align} Let $\ x ^*=A\sqrt[b-a]{\frac{a^2}{b^2}}\ $. Then $\ f(x)\ $ strictly decreases from $\ f(0)=-1\ $ down to $\ f(x^*)\ $ for $\ 0\le x\le x^*\ $, reaches a global minimum at $\ x=x^*\ $ and then strictly increases to $\ \infty\ $ for $\ x\ge x^*\ $. The equation $$ f(x)=v $$ therefore has no solutions for $\ v<f(x^*)\ $, exactly two solutions for $\ v\in(f(x^*),-1]\ $ and exactly one solution for $\ v\in\big\{f(x^*)\big\}\cup(-1,\infty)\ $. Thus, the equation $$ f(x)=0 $$ has exactly one root.