Graphing log with number in front of "log"

Question

When I have something like $y = log_2(x)$ I know that I have to turn it into exponential form and get: $2^y = x$. Next, I make a table for $X,Y$ and choose about 5 values for $y$, typically $-1, 0, 1, 2, 3$, then I plug it in and figure out the corresponding $x$ values.

This is simple, because once you have the table all you have to do is plot the points, however, what do I do when I have something like $y = -2log_2(x)$? Wouldn't that give me a $\pm$ scenario because that's the same as saying $y = log_2(x^2) = 2^y = x^2$?

I have no idea what to do from here.

Answer 1

Given $y=-2\log_2(x)$, you have $\frac{y}{-2}=\log_2(x)$ so $2^{-(y/2)}=x$. Now you can choose values of $y$ like you did before and plug them in to get $x$-values.

If you wanted to move the $-2$ into the log, you would get $y=\log_2(x^{-2})=\log_2\left(\frac{1}{x^2}\right)$, not $y=\log_2(x^2)$. Then $2^y=\frac{1}{x^2}$, so $x^2=\frac{1}{2^y}$. Now you're wondering whether you need to consider the positive and negative square root when you solve for $x$. The answer is no, because $\log_2(x)$ does not exist for negative values of $x$. The log function is only defined for positive inputs, so you graph will only contain points $(x,y)$ for which $x>0$.

Answer 2

That shouldn't give you a $\pm$ scenario, because the range of $x$ in a function $y=-2log_2(x)$ would be $(0,\infty)$. Therefore, even though you have $2^y=x^2$, you would want your x to be always positive, and that leads to a unique y value, so you would not actually deal with the $\pm$ scenario.