Natural log Graph

[MadSpaceCow]

Now, I am in Pre-cal, and I couldn't find a category for it, so I apologize if this is in the wrong area.

My teacher has given us a series of graphing problems to complete, but there is one that I could not figure out.

Graph f(x) = 3-ln(2x/(x+2))

I tried converting it to e^y-3 = 2x/(x+2) and plotting points but they didn't work right. I got an x/y Chart like so:

X |Y

2 |3
.0511 |0
1.2342|-6

Any help would be appreciated!

 

[HallsofIvy]

The simplest way to approach this, I think, is to use the laws of logarithms: y= 3- ln(2x/(x+2))= 3- ln(x)+ ln(x+2)- ln(2). Then it becomes fairly easy to find x, y values: x= 1, y= 3+ ln(3)- ln(2)= 3.405, x= 2, y= 3- ln(2)+ ln(4)- ln(2)= 3- ln(4)= 1.614, x= 3, y= 3- ln(3)+ ln(5)- ln(2)= 2.818, etc.

 

[MadSpaceCow]

Wow, thank you so very much! I never thought of that

 

[lookagain]

graph f(x) = 3-ln(2x/(x+2))

the simplest way to approach this, i think, is to use the laws of logarithms:
Y= 3- ln(2x/(x+2))= 3- ln(x)+ ln(x+2)- ln(2).​

These are not equivalent. In the original expression for f(x), x can be any negative
number less than -2. That is not so for the version of the alleged rewrite
by HallsofIvy, where x must be positive.


Consider f(x).

See that the argument for ln((2x)/(x + 2)), which is (2x)/(x + 2), must be positive.


When (2x)/(x + 2) > 0 is solved, the solution is x < -2 or x > 0.

Or, in interval notation, it is (-oo, -2) U (0, oo).

That is the domain of f(x).


-------------- ------------------


Also, for rough sketching, you can look to see/determine if there are any
vertical and/or horizontal asymptotes.



Vertical asymptote(s)
---------------------

2x cannot equal 0, and the limit of f(x) as x --> 0^+ equals oo, so
x = 0 is a vertical asymptote.


Also (x + 2) cannot equal 0. And the limit of f(x) as x --> -2^- equals -oo.
So then x = -2 is also a vertical asymptote.


. . . . . . . .. . . . . .


Horizontal asymptote(s)
-----------------------

As x --> oo, as well as when x --> -oo, f(x) approaches (3 - ln(2)), so the
horizontal asymptote is y = 3 - ln(2).

Or, y = approximately 2.3 (rounded to the nearest tenth)
​

 

[Deleted member 4993]

Now, I am in Pre-cal, and I couldn't find a category for it, so I apologize if this is in the wrong area.

My teacher has given us a series of graphing problems to complete, but there is one that I could not figure out.

Graph f(x) = 3-ln(2x/(x+2))

I tried converting it to e^y-3 = 2x/(x+2) and plotting points but they didn't work right. I got an x/y Chart like so:

X |Y

2 |3
.0511 |0
1.2342|-6

Any help would be appreciated!

Use wolframalpha.com - making sure you are looking at real-value plots.