# Log transformations question

1. Oct 15, 2009

### Matt1234

Here is what the graph looks like on a graphing calculator (notice the equation at the top):

http://img62.imageshack.us/img62/8898/graphingcalc.jpg [Broken]

Here is what my graph looks like:

http://img53.imageshack.us/img53/7475/lastscanc.jpg [Broken]

I dont understand why the asymptote is shifted 6 units left instead of 3 units left, And the point on my graph dont seem quite right.

i have the points (-1, -3)
Yet the graphing calculator has (-1, -3.79)

Note : where i wrote original points im referring to y = log x

I dont see what i hsve done wrong, can someone help me.

Thanks.

Last edited by a moderator: May 4, 2017
2. Oct 15, 2009

### Staff: Mentor

Your first step should have taken care of the compressions/expansions, so you should have looked at y = 2ln(1/2*x) first. The 1/2*x inside the parentheses causes an expansion away from the vertical axis. After that, take care of reflections, and finally translations.

3. Oct 15, 2009

### Matt1234

thanks will try that

Last edited: Oct 15, 2009
4. Oct 15, 2009

### Staff: Mentor

Notice that y = -2log(.5x + 3) + 3 = -2log(.5(x + 6)) - 3. This means that
1. The domain of this function is {x | x > -6}
2. The graph of y = -2log(.5x) has been translated to the left by 6 units, and down by 3 units.

5. Oct 15, 2009

### Matt1234

You hit the nail on the head my friend, I forgot to factor Stupid me!!!

6. Oct 15, 2009

### Matt1234

ill try again and post back my result

7. Oct 15, 2009

### Matt1234

Got it thank you very much, forgot to factor.

Here it is:
http://img62.imageshack.us/img62/2599/lastscanjy.jpg [Broken]

Last edited by a moderator: May 4, 2017