How many integer values of $y \in (-2022; 2022)$ are there such that there exists $x \in \mathbb R$ where the following condition is satisfied? $$2\log_2(x + y\sqrt{3}) - 2 = \log_{\sqrt{3}}(x^2 + y^2 - 1)$$
First of all, through the magic of throwing stuff around until it makes sense, we have that $$\begin{aligned} 2\log_2(x + y\sqrt{3}) - 2 = \log_{\sqrt{3}}(x^2 + y^2 - 1) &\iff \log_2(x + y\sqrt{3}) - 1 = \log_3(x^2 + y^2 - 1)\\ &\iff \log_2(x + y\sqrt{3}) = \log_3(3x^2 + 3y^2 - 3) \end{aligned}$$
[For context, this question is taken from an exam whose format consists of 50 multiple-choice questions with a time limit of 90 minutes. Calculators are the only electronic device allowed in the testing room. (You know those scientific calculators sold at stationery stores and sometimes bookstores? They are the goods.) I need a solution that works within these constraints. Thanks for your cooperation, as always. (Do I need to sound this professional?)
By the way, if the wording of the problem sounds rough, sorry for that. I'm not an expert at translating documents.]
The mathematical guacamole strikes back, and not in a good way...
Let $y\sqrt{3} = z$ and $x + z = t$, and the above becomes $$\log_2t = \log_3(3(t - z)^2 + z^2 - 3) \iff t^{\log_{2}{3}} = 3t^2 - 6zt + 4z^2 - 3$$
In order for the above equation to have a real root $z$, it must be met that $4t^{log_{2}{3}} - 3t^2 + 12 \ge 0$.
Let $f(t) = 4t^{log_{2}{3}} - 3t^2 + 12 \implies f'(t) = 4\log_{2}{3}t^{\log_{2}{3} - 1} - 6t$. The equation $f'(t) = 0$ has two roots $t = 0$ and $t = \left(\dfrac{2}{3}\log_{2}{3}\right)^{\log_{4/3}{2}} (= a)$, I guess. Look, my handheld gizmos aren't going to be able to accurately calculate the roots of the above equation, the same goes with $f(t) = 0$, but I can tell that it's very close, and I mean inconsequentially approximate to $t = 4$.
The table of variation of function $f(t)$ is as followed -
The roots of the equation $t^{\log_{2}{3}} = 3t^2 - 6zt + 4z^2 - 3$ are $z = \dfrac{3t \pm \sqrt{4t^{log_{2}{3}} - 3t^2 + 12}}{4}$.
I give up, （◎。◎；）, this is too difficult, and it's not even efficient.
The following is a graph of the equation $2\log_2(x + y\sqrt{3}) - 2 = \log_{\sqrt{3}}(x^2 + y^2 - 1)$—a sightly lopsided horseshoe.
It seems like there are two integer values of $y$ which satisfy the condition given in the problem, $y = 0$ and $y = 1$.
For $y = 0$, WolframAlpha gives the value of $x \approx 1,20277$. And for $y = 1$, WolframAlpha gives the value of $x \approx -0,625558$ and $x \approx 1,44197$, which can only be calculate numerically.
There's definitely some way to do this without relying so much on online tools, but my brain has already melted into a puddle of undetermined substance.
Anyhow, as always, thanks for reading, (and even more so if you could help), have a wonderful tomorrow, everyone~
By the way, the choices were $5, 2022, 2$ and $1010$.