Is there method to calculate the equation in pic?

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Only x_k is unknown. Does any method to solve this equation exist? I have tried use solve() in matlab to solve this equation. But failed after any strive.

$$\large{\frac{y_3-y_1}{y_2-y_1} =\frac{(e^{-\beta q_k})^{x_k}-(e^{-\beta \hat q_k})^{x_k}}{(e^{-\beta q_k})^{x_k}-(e^{-\beta \tilde q_k})^{x_k}}}$$

$$\beta=0.5,y_1\neq y_2\neq y_3, q_k\neq \hat q_k \neq \tilde q_k \in [0, 1] \text{ and } x_k\in(0,1).$$

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It certainly doesn't have an algebraic solution. Your equation can be written as solving for $x\in(0,1)$ the equation: $$\frac{u^x-v^x}{u^x-w^x}=C$$

where $u,v,w$ are distinct positive numbers and $C\neq 0,1$.

If $C> 1$ you can let $t=\frac{1}{C}$ and you get the equation: $$w^x = (1-t)u^x + t v^x$$

If $C<0$ you can let $t=\frac{1}{1-C}$ and you get the equation:

$$u^x = (1-t)w^x + tv^x$$

If $C\in (0,1)$ you can let $t=C$ and you get:

$$v^x = (1-t)u^x + tw^x$$

Because of these, we can see that (since $x$ is not negative) that:

  • If $C>1$ then $w$ must be between $u$ and $v$.
  • If $C\in (0,1)$ then $u$ must be between $w$ and $v$.
  • If $C< 0$ then $v$ must be between $u$ and $w$.

So you can always reduce to an equation of the form:

$$a^x = (1-t)b^x + tc^x$$

If you know your $q$ values are positive, we also know that $a,b,c\in(0,1)$.

where $a,b,c$ is some re-arrangement of $u,v,w$ and $t\in(0,1)$.

In particular, obviously $x=0$ is a solution if outside your range.

I think we know there is a solution to this with $x\in(0,1)$ under the following conditions:

$a$ is strictly between $b$ and $c$ and one of the following is true:

  • $a>(1-t)b + t c$ and $\log a < (1-t)\log b +t\log c$

or

  • $a<(1-t)b + tc$ and $\log a > (1-t)\log b + t\log c$

Basically, if $f(\alpha,\beta,\gamma)=\alpha-(1-t)\beta-t\gamma$ then when $f(a,b,c)$ and $f(\log a,\log b,\log c)$ are different signs, you can always find such an $x$. If $f(a,b,c)=0$ then I don't think there is a solution - the two solutions are $x=0,1$. If $f(\log a,\log b,\log c)=0$, I think you can use $f(\log^2 a,\log^2 b,\log^2 c)$ instead.