So, a bit of context first, I was studying how to construct the most powerful test using the Nyman-Pearson Lemma. The example went like this: given the normal distribution with known variance $\sigma^2$ and unknown mean $\mu$ a simple null hypothesis $H_0:\mu=\mu_0$ is tested against an alternative hypothesis $H_1: \mu=\mu_1, \mu_1 > \mu_0$.
I then proceeded with calculating the test statistic $T(\vec{x}) = \displaystyle\frac{ \displaystyle\prod\limits_{k=1}^{n} f(x_k;\mu_1) }{ \displaystyle\prod\limits_{k=1}^{n} f(x_k;\mu_0) } = {\large e}^{ \displaystyle\frac{n\overline{x}}{\sigma^2}(\mu_1 - \mu_0) - \displaystyle\frac{n}{2\sigma^2}(\mu_1^2 - \mu_0^2) } $
After that, following the definition of the critical region $K_\alpha = \left\{\vec{x} = (x_1, ..., x_n): \quad T(\vec{x}) \gt c_\alpha \right\}$, I solved a following inequality for $\overline{x}$: $${\large e} ^{ \displaystyle\frac{n\overline{x}}{\sigma^2}(\mu_1 - \mu_0) - \displaystyle\frac{n}{2\sigma^2}(\mu_1^2 - \mu_0^2) } \gt c$$ $$\overline{x} \gt \displaystyle\frac{\sigma^2}{n(\mu_1 - \mu_0)}\ln c + \displaystyle\frac{\mu_1 + \mu_0}{2}$$
The example mentioned how before we used a different critical region to test such hypotheses: $K_\alpha = \left\{\vec{x} = (x_1, ..., x_n): \quad \overline{x} \gt c_\alpha, \quad c_\alpha = \mu_0 + \displaystyle\frac{\sigma}{\sqrt{n}}z_\alpha \right\}$
And then it was suggested to show that these 2 critical regions are equivalent by solving a following equation:
$$\displaystyle\frac{\sigma^2}{n(\mu_1 - \mu_0)}\ln c + \displaystyle\frac{\mu_1 + \mu_0}{2} = c_\alpha = \mu_0 + \displaystyle\frac{\sigma}{\sqrt{n}}z_\alpha$$
When solved for $c$, it looks like this: $$c = e^{\displaystyle-(\mu_1 - \mu_0) \left[ \displaystyle\frac{n}{2\sigma^2} (\mu_1 - \mu_0) - \displaystyle\frac{\sqrt{n}}{\sigma} z_\alpha\right]}$$
And immediately after that a conclusion is made that these 2 critical regions are equivalent.
And now my question:
I might be missing something stupidly obvious here, but I'm having doubts about how to interpret this result. Is it that there exists such $c$ that satisfies the equation? Or is there a different reason why this conclusion can be made?