How can I solve for r in the equation 2x = x^r for 0<x<1?
I did the following:
log 2 + log x = r log x
r = log 2 / log x + 1
However, for the value of x = .5, r evaluates to 0, which gives you the equation 2x = 1.
what is wrong with $\displaystyle r = 0$ ? $\displaystyle x$ is the variable that has the restriction.
$\displaystyle 2x - x^r = 0$
$\displaystyle x(2 - x^{r-1}) = 0$
$\displaystyle x = 0$ ... invalid solution since $\displaystyle 0 < x < 1$
$\displaystyle x^{r-1} = 2$
$\displaystyle \displaystyle x = 2^{\frac{1}{r-1}}$
since $\displaystyle 0 < x < 1$ ...
$\displaystyle \displaystyle \frac{1}{r-1} < 0$
$\displaystyle r < 1$