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Given:

$\ln(xa)= b\ln(c-x)$

I am unsure of how to manipulate the values within the natural logs to solve for x while the factor b remains. I can safely move in circles by applying the definition of the logarithm to yield the exponential form.

$ax = (c-x)^{b}$

Is there a way to make forward progress? I know all values ($a$, $b$, $c$, $x$) to be real and $b$ to be a positive integer. I am only interested in real solutions.

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    Can you do this special case: $x = (1-x)^{10}$. If you cannot do that, there is not much hope for the general one.2012-02-18

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Rewriting your last expression as $a=\dfrac{(c-x)^b}{x},$ you can see that, for positive values of $a$ (or all nonzero values for odd $b$), this is $a^{1/b}=\dfrac{c-x}{x^{1/b}},$ which at least gets you to a polynomial (changing $x^{1/b}$ to $x$): $x^b+a^{1/b}x-c=0$. I do not know of a simple way to characterise the solutions of such an equation. I find this polynomial more pleasing than the one you'd get from the binomial theorem.

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    You could let $y=c-x$ in the original equation and get the equally pleasing $y^b + ay - ac = 0$ without any extra conditions.2012-02-18