I have encountered this equation:
$$x^y = ax-b$$
I know to find $y$ as a function of $x$ then:
$$y\ln(x) = \ln(ax-b)$$
$$y = \frac{\ln(ax-b)}{\ln(x)}$$
or
$$y = \log_x(ax-b)$$
But the problem I need to find $x$ as a function of $y$ and don't know how .
If $y\in \Bbb N$, then the problem is finding roots of a polynomial of degree $y$. This problem has general closed-form representations of roots only for $y\in\{1,2,3,4\}$, for example solutions for linear, quadratic, cubic and quartic equations for the cases 1, 2, 3 and 4, respectively.
Apart from that, there are only few cases that can be reduced to the solutions above:
$y\in\{-1,-2,-3\}$: Multiply the equation with $x^{-y}$ will result in solving a polynomial equation of degree $y+1$. Notice that $x=0$ needs special treatment.
$y\in\{\frac12,\frac13,\frac14\}$: Substituting $x=z^{1/y}$ will yield a polynomial of degree $1/y$ in $z$.
$y\in\{-\frac12,-\frac13\}$: Substituting $x=z^{-1/y}$ will get us to a polynomial of degree $1-1/y\in{3,4}$.
Apart from that, there are numerical methods like Newton-Raphson iteration that converge very quickly (quadratically) in the general case:
Approximating zeros of $f(x) = x^y-ax+b=0$ by picking some initial value and then iterating
$$x\mapsto x-\frac{f(x)}{f'(x)} = x-\frac{x^y-ax+b}{yx^{y-1}-a} = \frac{(y-1)x^y-b}{yx^{y-1}-a} $$
The closest you can get to finding $x$ as a function of $y$, is using the implicit function theorem around a solution of the equation. So let $g(x,y) = x^y-ax+b = 0$ then the total differential is
$$(yx^{y-1}-a)dx + x^y\ln x \,dy = 0$$
so that
$$\frac{dx}{dy} = \frac{x^y\ln x}{a-yx^{y-1}}=:h(x,y)$$
This means that if you found a solution $g(x_0,y_0) = 0$, and you change $y$ a little bit to $y_0+\Delta y$, then the adjusted solution for $x_0$ is around
$$x_0+\Delta x \approx x_0 + \Delta y\frac{x_0^{y_0}\ln x_0}{a-y_0x_0^{y_0-1}} =\Delta y\cdot h(x_0,y_0) $$
This is only a hint for $x>0$ (so that we can take $\ln x$) and where the denominator of $h$ is not zero.
Say we have $a=2$ and $b=-4$, then $(x,y) = (2,3)$ is a solution to $g(x,y)=0$. We want to estimate $x$ when we change $y$ slighly to $y+\Delta y=3.01$. The estimate above tells us we have to expect $x+\Delta x\approx 1.9944548$ whereas the actual solution is around $1.9944853$.
Here are $2$ solutions using Wolfram Alpha with a similar derivation to this solution. The function used is Inverse Beta Regularized $\text I^{-1}_s(a,b)$ which is usually a quantile function. To use it, one must have $0\le s\le 1;a,b>0$:
$$x^y-ax+b=0\implies x=\frac{by\,\text I^{-1}_{-\frac{b^{y-1}(1-y)^{1-y}y^y}{(-a)^y}}(-y,2)}{a(y-1)},y<0,0\le -\frac{b^{y-1}(1-y)^{1-y}y^y}{(-a)^y} \le 1$$
Test the left formula here
and
$$x^y-ax+b=0\implies x=\frac{by}{a(y-1)\text I^{-1}_\frac{b^{y-1}(y-1)}{(-a)^y\left(\frac1y-1\right)^y}(y-1,2)},y>1,0< \frac{b^{y-1}(y-1)}{(-a)^y\left(\frac1y-1\right)^y} <1$$
Test this formula here. For $0<y<1$, try the substitution of $x\to x^{\frac1y}$ and cancel powers.
The question’s creator also mentioned excel in a comment, so try using the BETA.INV function instead of InverseBetaRegularized
We have
$$y=-\frac{\ln(ax-b)+2k\pi i}{\ln(x)}.$$
The function $x\mapsto -\frac{\ln(ax-b)+2k\pi i}{\ln(x)}$ doesn't have a partial inverse that is an elementary function.
It's an open question if the equation has solutions that are elementary numbers.
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$$x^y=ax-b$$
$$x^y-ax+b=0$$
$$e^{\ln(x)y}-ax+b=0$$
LambertW cannot be applied either.
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$$x^y=ax-b$$
For rational $y$, this equation is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.
For rational $y\neq 0,1$, the equation is related to a trinomial equation.
For real or complex $y\neq 0,1$, the equation is in a form similar to a trinomial equation. A closed-form solution can be obtained using confluent Fox-Wright Function $\ _1\Psi_1$ therefore.
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