# Math Help - Exponential equation

1. ## Exponential equation

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.

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 $r = 0$ ? $x$ is the variable that has the restriction.

$2x - x^r = 0$

$x(2 - x^{r-1}) = 0$

$x = 0$ ... invalid solution since $0 < x < 1$

$x^{r-1} = 2$

$\displaystyle x = 2^{\frac{1}{r-1}}$

since $0 < x < 1$ ...

$\displaystyle \frac{1}{r-1} < 0$

$r < 1$

3. Thanks a lot, skeeter.