#### Dr.Peterson

##### Elite Member

- Joined
- Nov 12, 2017

- Messages
- 4,110

I agree with most of this.I don't understand "log_{2}times y", but I do have some final comments on this exercise. In the op, you had made it sound like this was a graphing exercise. But, it's not.They gave you the graph.

I had thought it awkward for an exercise to ask students to graph x=2∙log_{2}(y-2) because that's a logarithmic function except that the independent and dependent variables have been switched. (I wasn't sure who switched the variables, and I've seen instructors do some weird presentations with coordinate axes.) This is why I asked to see the complete exercise statement (you provided it in post #12). This is also why theforum guidelinesask students to post exercises as given. Once I saw the actual exercise, the axis labels made clear that x=2∙log_{2}(y-2) represents the inverse of function y=2∙log_{2}(x-2).

Others may disagree, but to me the exercise is not asking you to find coordinates of points or to draw a graph.I think the intent is for students to use attributes they were taught regarding shapes of logarithmic and exponential curves, graph shifts (i.e., all inputs or outputs changed by a constant) and the reflective graph symmetry about the line y=x between a curve and its inverse. In other words, this exercise can be answered without doing any numerical calculations.

The graph of y=2∙log_{2}(x-2) is the graph of y=2∙log_{2}(x) shifted two units to the right. In other words, we know the line x=2 is a vertical asymptote because we've learned the general shape of logarithmic function graphs. We reflect the shifted graph across the line y=x, to identify the inverse graph x=2∙log_{2}(y-2). Everything gets reflected across the line y=x -- including the asymptote. That is, the vertical asymptote x=2 becomes the horizontal asymptote y=2 on the inverse graph,andthe exponential curve must be increasing over its domain. So, we can eliminates choices A, B and C right away -- without solving for y or finding points.

As you can infer from my post #3, the picture of the problem (reflections and all) was there originally (though I found it unreadable); at some point it disappeared, and was reintroduced later (post #12). It does explicitly say to "sketch the graph", so at some level it

*is*a graphing exercise; but as a multiple choice, it doesn't require actually doing that, but emphasizes the direction of the curve, which is all about the transformations (reflections in particular).

I agree that it is surely intended to test transformations; apart from the visual aspect of the inverse (which I like), what you say agrees pretty much with my suggestions back in post #3.

At any rate, it's a hard problem for a student who hasn't quite grasped either the transformations or the meaning of the log. and maybe it's accomplished its goals. But there's still a lot to discuss with ironsheep.