This is not a weird question at all, but I don't think this is the best forum for a lesson on logarithms. You'll find lots of stuff on the Web if you Google "logarithm." I suggest you look at some of the lessons, and come back here if you have questions. In the meantime, I'll explain how to answer the question.
What you want is the base-$2$ logarithm. For example $$\log_2 7=x$$ means the same thing as $$2^x=7.$$ Now, to compute $\log_2{7}$ we can use WolframAlpha and find $$\log_2{7}\approx2.8$$ This tells us that the $n$ we seek is either $2$ or $3$. While it's easy to guess that the answer is $3$ in this case, until you know more about logarithms, I suggest you just compute bot $2^2$ and $2^3$ and see which is closer. It's not always true that you can just take the integer closest to the logarithm.
Hope this helps.
EDIT
Let's say that $t=\left\lfloor\log_2 n\right\rfloor$, that is, $t$ is the greatest integer less than or equal to $\log_2 n$, so we know $$2^t\leq n< 2^{t+1}$$
We want to set $m=t$ if $n\leq \frac32\cdot2^t$ that is, if $$\log_2 n<\log_2\frac32+t=t+.584963...$$
so you compute $\log_2 n$, and round down if the integer part is less than $.584963$, otherwise, round up.
If $n$ happens to be very close to $3\cdot2^m$ for some integer $m$, you may still have to compute $2^m$ and $2^{m+1}$ to be sure which to take, because of roundoff error in computing the logarithm.
Best Answer
There will be infinitely many pairs $(x,y)$ of real numbers that satisfy the equation. If you are looking for integer solutions, that is another matter.
I would rewrite the equation as $9xy+42x+51y+213=0$, and then as $$(3x+17)(3y+14)-(17)(14)+213=0,$$ which turns into the attractive $$(3x+17)(3y+14)=25.$$ Note that we did an analogue of completing the square. We get a hyperbola.
Now for integer solutions the analysis becomes quite simple. We take all ordered pairs $(s,t)$ of integers (both positive or both negative) such that $st=25$. There are only $6$ such pairs.
For some but not all of these pairs, it turns out that $x$ and $y$ are integers. Let's start. Look at $s=1, t=25$. No good, there is no integer $x$ such that $3x+17=1$. Look at $s=-1,t=-25$. That gives $3x+17=-1$, $3y+14=-25$, which has the integer solution $x=-6, y=-13$. Continue. There is not far to go!