I'm looking for a closed form solution for $a^x - b^x = 1$ where $a,b\in \mathbb{R}$ are known and $x$ is unknown. There is a similar problem with positive sign instead of minus. Here are my attempts:
Let $b = a+\epsilon $ where $\epsilon = b-a$ and use Newton's generalized binomial theorem to get $$\sum_{i=1}^\infty\binom{x}{i}\epsilon^ia^{x-i} = -1$$but I don't know how to proceed. Also I think for $a\gt b \gt 0$ there is a unique solution and for $0\lt a\lt b$ there is no solution since $a^x$ is an increasing function. We can rewrite the equation as $$a^{-x}(1+b^x) = 1 \implies -x\ln a + \ln(1+b^x) = 0 \implies x =\frac{\ln(1+b^x)}{\ln a}$$Assuming that we can use Taylor series for $\ln(1+b^x)$, we have $$x = \frac{1}{\ln a}(b^x-\frac{b^{2x}}{2}+\frac{b^{3x}}{3}-\frac{b^{4x}}{4}+\cdots)$$and this seems useless.
Add $b^x$ to get $$a^x = 1+b^x$$ then divide by $a^{x}$ to get $$1 = \left(\frac{1}{a}\right)^x+\left(\frac{b}{a}\right)^x$$
Assuming that $a, b>0$, if $\frac{1}{a}, \frac{b}{a}<1$, there is a solution. If only one out of $\frac{1}{a}, \frac{b}{a}$ is less than $1$, there will be no solution, and if both are greater than $1$, the solution will be the negative of the solution for $1=a^x + \left(\frac{a}{b}\right)^x$.
This question is now the same as the linked question. Let's say we are trying to solve $$1=a^x + \left(\frac{a}{b}\right)^x$$
Using this answer, if $b < 1$, let $p = \frac{\ln(a/b)}{\ln(a)}$ and $u = a^x$. Otherwise, let $p = \frac{\ln(a)}{\ln(a/b)}$ and $u = \left(\frac{a}{b}\right)^x$. Then $$u = \sum_{k=0}^\infty \frac{\Gamma(pk+1)(-1)^k}{\Gamma((p-1)k+2) k!}$$