equation $a^x+b^x-c^x = 1$ Is there any way to solve analytically an equation for $x$:
\begin{equation}
a^x+b^x-c^x = 1,
\end{equation}
where $a$, $b$, and $c$ are arbitrary real numbers? If it helps, $a$ is about 6, $b$ is about 2, and $c$ is about 7.
If not, is there a way to at least get an approximate value of $x$?
 A: An obvious solution is $x=0$.
If $a=6$ , $b=2$ , $c=7\quad$ a second solution is $x=1$.
If $a\simeq 6$ , $b\simeq 2$ , $c\simeq 7\quad$ the shape of the curve
$$y(x)=a^x+b^x-c^x$$
is of this kind :

The second solution is close to $x\simeq 1$ and has to be computed thanks to numerical calculus with given numerical values of $a,b,c.$
A first approximate can be computed thanks to serie expansion of :
$$a=6+\epsilon_1$$
$$b=2+\epsilon_2$$
$$c=7+\epsilon_3$$
$$x=1+\epsilon$$
From the fist term of the series expension of
$$(6+\epsilon_1)^{1+\epsilon}+(2+\epsilon_2)^{1+\epsilon}-(7+\epsilon_3)^{1+\epsilon}=1$$
one get :
$$x\simeq 1-\frac{\epsilon_1+\epsilon_2-\epsilon_3}{6\ln(6)+2\ln(2)-7\ln(7)}$$
For example with $a=6.2$ , $b=1.9$ , $c=7.3$ :
$$x\simeq 0.865276\quad;\quad a^x+b^x+c^x\simeq 1.006577$$
Comparing to the solution obtained with numerical nonlinear equation solving (Wolfram Alpha) :
$$x\simeq 0.870988\quad;\quad a^x+b^x+c^x\simeq 0.9999997$$
DETAILS OF THE SERIES EXPANSIONS (answer to a request in comments).
$$(6+\epsilon_1)^{1+\epsilon}=\exp\big((1+\epsilon)(\ln(6+\epsilon_1) \big)$$
$$(1+\epsilon)\ln(6+\epsilon_1)=(1+\epsilon)\left(\ln(6)+\ln(1+\frac{\epsilon_1}{6})\right)$$
In order to simplify the writting, the terms of second order such as $(\epsilon_1)^2$ or such as $(\epsilon_1\epsilon)$ are neglected. Each time they are replaced by ...
$$(1+\epsilon)\ln(6+\epsilon_1) \simeq (1+\epsilon)\left(\ln(6)+\frac{\epsilon_1}{6}+...\right)$$
$$(1+\epsilon)\ln(6+\epsilon_1) \simeq \ln(6)+\frac{\epsilon_1}{6} +\epsilon \ln(6)+...$$
$$\exp\big((1+\epsilon)\ln(6+\epsilon_1)\big) \simeq 6\exp\left(\frac{\epsilon_1}{6} +\epsilon \ln(6)+... \right)$$
$$\exp\big((1+\epsilon)\ln(6+\epsilon_1)\big) \simeq 6\left(1+\frac{\epsilon_1}{6} +\epsilon \ln(6)\right)+...$$
$$\exp\big((1+\epsilon)\ln(6+\epsilon_1)\big) \simeq 6+\epsilon_1 +6\,\epsilon \ln(6)+...$$
The same for the other terms :
$$\exp\big((1+\epsilon)\ln(2+\epsilon_2)\big) \simeq 2+\epsilon_2 +2\,\epsilon \ln(2)+...$$
$$\exp\big((1+\epsilon)\ln(7+\epsilon_3)\big) \simeq 7+\epsilon_3 +7\,\epsilon \ln(7)+...$$
Putting them into
$$(6+\epsilon_1)^{1+\epsilon}+(2+\epsilon_2)^{1+\epsilon}-(7+\epsilon_3)^{1+\epsilon}=1$$
leads to
$$6+\epsilon_1 +6\,\epsilon \ln(6)+2+\epsilon_2 +2\,\epsilon \ln(2)-7-\epsilon_3 -7\,\epsilon \ln(7)+...\simeq 1$$
$$\epsilon_1 +6\,\epsilon \ln(6)+\epsilon_2 +2\,\epsilon \ln(2)-\epsilon_3 -7\,\epsilon \ln(7)+...\simeq 0$$
Solving for $\epsilon$ :
$$\epsilon\simeq-\frac{\epsilon_1+\epsilon_2-\epsilon_3}{6\ln(6)+2\ln(2)-7\ln(7)}$$
A: In the spirit as my friend @JJacquelin, let $x=1+t$ and consider the function
$$f(t)=(6+\alpha )^{t+1}+(2+\beta )^{t+1}-(7+\gamma)^{t+1}-1$$ and expand it as a series around $t=0$
$$f(t)=\sum_{n=0}^\infty  {A_n}\, t^n$$ where
$$A_0=\alpha+\beta-\gamma$$
$$A_n=\frac 1{n!}\Big[(6+\alpha )\big[\log(6+\alpha )\big]^n+(2+\beta )\big[\log(2+\beta )\big]^n-(7+\gamma)\big[\log(7+\gamma )\big]^n\Big]$$ Truncate it to any order and use series reversion.
Using  three terms only, we should have
$$t\sim -\frac{A_0 }{A_1 }-\frac{A_0^2\,A_2 }{A_1^3 }-\frac{A_0^3\,(2A_2^2 -A_1\,A_3)}{A_1^5 }$$
For the same example where $a=6.2$ , $b=1.9$ , $c=7.3$, the above would give
$$t=-0.125744 \quad \implies \quad x=0.874256  \quad \implies \quad a^x+b^x-c^x=0.99616$$
Using twice more terms in the expansion
$$t=-0.128767 \quad \implies \quad x=0.871233  \quad \implies \quad a^x+b^x-c^x=0.99971$$
