If $x-x^p = 1$ then how can I define x without itself?

Question

I have this equation which shows the relation between $p$ & $x$. Here both $p$ and $x$ are real numbers.

$x-x^p = 1$

When I know $x$, I can find $p$ using $p=\frac{ln(x-1)}{ln(x)}$

Now my question is, how can I find a solution for $x$ when I know $p$ (similar to the logarithmic solution of $p$)?

Answer 1

I assume you are interested in positive real solutions?

If we consider $x=1+x^p$ we see that it depends on $p$ whether there exists a positive real solution or not. If $p\geq 1$ you have no real solutions, since $x\in[0,1)$ implies $x< 1\leq 1+x^p$ and $x\geq 1$ implies $x\leq x^p<1+x^p$.

If $p<1$ we define $g(x)=1+x^p-x$. We see $\lim_{x\to 0^+}g(x)\geq 1$ and $\lim_{x\to\infty} g(x)=-\infty$. The IVT yields a root of $g$ in $(0,\infty)$.

For a general $p$, you will need a numerical approximation for the root of $g$.

Answer 2

This is not a solution, it is only my Idea. I will be glade if anyone could be able to complete it.

If there is a function $f(p)$, such that $x=f(p)$, then clearly it must be continious, at least on positives real numbers.

At first let $p$ to be a rational number. i.e. $p=\dfrac{m}{n}$. So $x$ will be a root of:

$(x-1)^n=x^m$.

If we are able to prove that there is no such a function for almost all rational numbers $p=\dfrac{m}{n}$, then it follows that there is no such a function for reals.