Log-Plots: Sketching

  • #1
479
12

Homework Statement



Screen Shot 2017-08-19 at 10.35.02 PM.png


Homework Equations




The Attempt at a Solution


I have been trying to rearrange the equation to look something like ln(y) = Aln(x) but to no avail. Something tells me that that's not necessarily the right way to approach the problem, could anyone point out what I should be trying to do? Thanks!
 

Answers and Replies

  • #2
34,474
10,596
The graph won't be a straight line on the log-log-plot.

You can find individual points, and then connect them with a line.
 
  • #3
20,315
4,300
What does ##\ln{(x^2 +2x+1)}## approach for small values of x (linear approximation in x)? What does ##\ln{(x^2 +2x+1)}## approach at large values of x (linear in ##\ln{x}##)? Plot these on your graph with dashed lines to indicate asymptotes.
 
  • #4
479
12
Thanks for the replies.

Does this mean i should be plotting lny = ln(x2+2x+1) instead of trying to resolve ln(x2+2x+1) into something like lnx?
 
  • #5
20,315
4,300
Thanks for the replies.

Does this mean i should be plotting lny = ln(x2+2x+1) instead of trying to resolve ln(x2+2x+1) into something like lnx?
Read my post again.
 
  • #6
479
12
Read my post again.
So i've done a bit of a sketch and ln(y) dips to negative infinity (asymptote) at x = -1, it then rises slowly to infinity from x = ±1 to infinity axis (symmetrical about x = -1). I don't know how to go on from here, any further pointers are greatly appreciated.
 
  • #7
20,315
4,300
At x = 1, ln(y) = ln(4).

At very large x, ln(y) approaches ##\ln(x^2)=2\ln{x}##

What does y approach at small x (i.e., linear approximation)?
 
  • #8
34,474
10,596
So i've done a bit of a sketch and ln(y) dips to negative infinity (asymptote) at x = -1, it then rises slowly to infinity from x = ±1 to infinity axis (symmetrical about x = -1). I don't know how to go on from here, any further pointers are greatly appreciated.
There are no negative x-values in your logarithmic plot. What is y at x=0? What about very small x-values (e. g. 10-100)?
 
  • #9
479
12
There are no negative x-values in your logarithmic plot. What is y at x=0? What about very small x-values (e. g. 10-100)?
When x goes to 0, y tends to x0 and ln(y) ≈ 0ln(x)?

Apologies if this is really obvious, I'm having a hard time grasping this. Thanks for your patience
 
  • #10
20,315
4,300
For small values of x, y approaches 1+2x, so ln(y) approaches ##2x=2e^{\ln{x}}##
 
  • #11
34,474
10,596
When x goes to 0, y tends to x0
Why x0? The answer is a real number, it shouldn't have any x in it.

@Chestermiller: That is not the point of my question. A much easier estimate is needed here.
 
  • #13
vela
Staff Emeritus
Science Advisor
Homework Helper
Education Advisor
14,659
1,292
I have been trying to rearrange the equation to look something like ln(y) = Aln(x) but to no avail. Something tells me that that's not necessarily the right way to approach the problem, could anyone point out what I should be trying to do? Thanks!
You might find it helpful to rewrite the function slightly as ##y = (x+1)^2## so that ##\ln y = 2 \ln (x+1)##.

On the horizontal axis, you have ##-\infty < \ln x < \infty## which means ##0 < x < \infty##. ##\ln x = 0## corresponds to ##x=1##, so the left half of the axis corresponds to the range ##0 < x < 1## and the right half of the axis, to ##x > 1##.

What is ##\ln y## approximately equal to when ##x\ll 1## (left end of the plot) and when ##x \gg 1## (right end of the plot)?
 
  • Like
Likes Chestermiller
  • #15
20,315
4,300
You don't need the linear term for this asymptote.
Sure you do. The asymptote I'm referring to is $$\ln{y}=2e^{\ln{x}}$$It's not linear or as simple as ##\ln{y}=0##, but it's much more accurate for x < 1 (and not too difficult to plot).
 
  • #16
34,474
10,596
What you want to study is not a straight line. Let OP start with a simpler problem before we go to more advanced steps.
 
  • #17
479
12
Would I be right in saying:

set x into a table, ie x = 0, x = 1 and so on
set corresponding y into a table ie y = 1, y = 4 and so on

take values of lnx and lny, and plot them onto the lnx and lny axes (ignoring x = 0 and y = 1).

Asymptote as x tends to ∞ is the line ln(y) = 2ln(x) due to x2 being the dominant term as x is nearing ∞.

Asymptote as x tends to 0 is the line ln(y) = ln(1).

However, I still don't quite understand what @Chestermiller is saying regarding the asymptote as x -> 0.

Thank you both for your patience.
 
  • #18
34,474
10,596
Asymptote as x tends to ∞ is the line ln(y) = 2ln(x) due to x2 being the dominant term as x is nearing ∞.

Asymptote as x tends to 0 is the line ln(y) = ln(1).
Right.
 
  • #19
20,315
4,300
Would I be right in saying:

set x into a table, ie x = 0, x = 1 and so on
set corresponding y into a table ie y = 1, y = 4 and so on

take values of lnx and lny, and plot them onto the lnx and lny axes (ignoring x = 0 and y = 1).

Asymptote as x tends to ∞ is the line ln(y) = 2ln(x) due to x2 being the dominant term as x is nearing ∞.

Asymptote as x tends to 0 is the line ln(y) = ln(1).

However, I still don't quite understand what @Chestermiller is saying regarding the asymptote as x -> 0.

Thank you both for your patience.
There is a second asymptote as x tends to 0 that provides a much closer approximation to the desired function than simply ln(y)=0. That asymptote is $$\ln{y}=2e^{\ln{x}}$$. Plot it up and see what you get in comparison to the desired function.
 
  • #20
34,474
10,596
Both functions in the relevant range. I don't see the point.

ln(y)=0 is an asymptote, a nice straight line that provides an excellent approximation for small x. What you suggest is an arbitrary function that is not a straight line and not the full function either.
 
  • #22
34,474
10,596
That is not the x-range where you would use an asymptote. It is between small x and large x.
 
  • #24
SammyS
Staff Emeritus
Science Advisor
Homework Helper
Gold Member
11,315
1,006
  • Like
Likes Chestermiller

Related Threads on Log-Plots: Sketching

  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
1
Views
2K
Replies
3
Views
1K
Replies
17
Views
6K
  • Last Post
Replies
2
Views
4K
  • Last Post
Replies
4
Views
2K
  • Last Post
Replies
3
Views
1K
  • Last Post
Replies
1
Views
1K
  • Last Post
Replies
10
Views
2K
Top