How many integer values of $y$ are there such that for each value of $y$, there are no more than $2021$ integer values of $x$ satisfying the condition that $\log_2(x + y^2 + 1) - 3^{y^2 + y - 3x} < 0$?
[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.]
Oftentimes, there are questions which I am afraid to ask, due to the fact that I literally don't know how to approach them. This is no exception. Why do these symbols seem like they're hastily chopped in the food processor? Why does this amalgamation of mathematical guacamole exist? ლ(ಠロಠ ლ)
Below is a graph from Desmos from the results for $\log_2(x + y^2 + 1) - 3^{y^2 + y - 3x} < 0$.
Plugging the equation $\dfrac{1}{(x + y^2 + 1)\ln 2} - 3^{y^2 + y - 3x + 1}\ln 3 = 0$, which is $\dfrac{\mathrm d}{\mathrm dx}\left[\log_2(x + y^2 + 1) - 3^{y^2 + y - 3x}\right]$ into WolframAlpha, and we get $x = -y^2 - \dfrac{W\left(\dfrac{1}{3^{4y^2 + y + 3}\ln 2}\right)}{3\ln 3} - 1$, which doesn't mean anything for me, unfortunately.
As always, thanks for reading, (and even more so if you could help, specifically with a solution which could be thought of in the time required), have a wonderful tomorrow, everyone~
Anyway, the choices were $110, 111, 109$ and $108$.
Actually, plugging $\log_2\left(x + \left(\dfrac{217}{4}\right)^2 + 1\right) - 3^{(217/4)^2 \pm 217/4 - 3x} < 0$ into WolframAlpha and we get $(-2944,0625; 998,352)$ and $(-2944,0625; 962,185)$, and there are $3943$ and $3907$ integers in those intervals respectively, so way offff~
Let's look at this as $$ y^2 + y -3x > \log_3\log_2(x+y^2+1) $$ This will obviously be satisfied for any $x < 0$ that doesn't cause the argument of the logarithm to become nonpositive. Thus we immediately have $y^2$ solutions for $-y^2 < x < -1$. For nonnegative $x$, note that $\log_3\log_2$ is an incredibly slowly growing function. In fact, we have $2 < \log_3\log_2 z < 3$ for all $512 < z < 10^8$. If we assume that having $2021$ solutions requires $y > 22$ (which it does), we can replace the inequality with $$ y^2 + y -3x \ge 3 $$ for nonnegative $x$. The number of solutions to this is easily calculated: $\lfloor(y^2+y)/3\rfloor$. So the total number of solutions for a given $y$ is $$ N(y) = \left\lfloor\frac{y^2+y}{3}\right\rfloor + y^2 $$ Now we just have to find the range of $y$ that satisfy $N(y) < 2021$. Noting that this should be somewhere near $\pm\sqrt{3\cdot2021/4}\approx 38.9$, we can just check $\pm 38$ and $\pm 39$ to see which work. This gives the range $y\in [-39,38]$, which is $78$ distinct values.
...which isn't one of the solutions. This WolframAlpha link calculates the number of solutions and seems to agree with me. On the other hand, if you don't allow negative $x$, you get $y\in [-78,77]$, which is $156$ distinct solutions, which is still not one of the answers. Very confusing. Are you sure you have everything correct about the problem statement?