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

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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$\ln(a*b)=\ln a+\ln b$ is definitively handy here. – Lazar Ljubenović Feb 18 '12 at 0:47
Can you do this special case: $x = (1-x)^{10}$. If you cannot do that, there is not much hope for the general one. – GEdgar Feb 18 '12 at 2:12

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. – Rahul Feb 18 '12 at 2:20

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