Solving $-1=e^a-2e^{av}$ as part of an equation system Problem
Given $f_2(x)=e^{ax-b}+c$ with $x \in \left(0,1\right)$, I am trying to calculate the parameters $a,b,c$ in respect to the following constraints:
$$
\begin{align}
  f_2(0) &= 0 \\
  f_2(1) &= 1 \\
  f_2(v) &= \frac{1}{2}
\end{align}
$$
Where $v \in \left(0.5,1\right)$ is a fixed parameter.
Current Situation
I have tried solving this analytically and have reached several other representations, including $-1=e^a-2e^{av}$, which should lead to a solution for $a$ but have not found one.
Wolfram-alpha can calculate one and given $v=0.6$ as an example, a solution is $a \approx 0.822163$, $b \approx 0.24373$, $c \approx -0.784057$ (see graph below).
Substitution of $e^a=u$ yields $-1 = u - 2u^v$ which seems doable ($v$ is constant), but I can't seem to find a solution myself.
Question
I would like to hear pointers as to how (and if) this can be solved analytically.
Context
I am trying to find a monotonic and continuous function $f$ that satisfies the above constraints in order to stretch the interval $\left(0,1\right)$ to have it's new center at $v$.
Such a function exists for the case $v \in \left(0,0.5\right)$ using the logarithm:
$$f_1(x) = a \cdot \ln(x+b)+c$$
The parameters can be determined analytically and given that $\exp$ is the inverse of $\ln$, I expect a similiar solution for $v > 0.5$ using $\exp$.

 A: 1)
$$f_2(x)=e^{ax-b}+c$$
Your constraints yield the equation system
$$\left\{e^{-b}+c=0,e^{a-b}+c=1,e^{av-b}+c=\frac{1}{2}\right\}.$$
This equation system has solutions if
$$b=\ln(e^{a}-1)+2k_1\pi i\ \land\ c=-(e^a-1)^{-1}\ \land\ v=\frac{1}{a}\left(\ln(\frac{1}{2}e^a+\frac{1}{2})+2k_2\pi i\right)$$
for $k_1,k_2\in\mathbb{Z}$.
2)
$$-1=e^a-2e^{av}$$
$$-1=u-2u^v$$
For rational $v$, this equation is related to an algebraic equation and we can use the known solution formulas and methods for algebraic equations.
For rational $v\neq 0,1$, the equation is related to a trinomial equation.
For real or complex $v\neq 0,1$, 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.
$\ $
Szabó, P. G.: On the roots of the trinomial equation. Centr. Eur. J. Operat. Res. 18 (2010) (1) 97-104
Belkić, D.: All the trinomial roots, their powers and logarithms from the Lambert series, Bell polynomials and Fox–Wright function: illustration for genome multiplicity in survival of irradiated cells. J. Math. Chem. 57 (2019) 59-106
