# Log-Plots: Sketching

## 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!

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mfb
Mentor
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.

Chestermiller
Mentor
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.

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?

Chestermiller
Mentor
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?

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.

Chestermiller
Mentor
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)?

mfb
Mentor
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)?

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

Chestermiller
Mentor
For small values of x, y approaches 1+2x, so ln(y) approaches ##2x=2e^{\ln{x}}##

mfb
Mentor
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.

Chestermiller
Mentor
That would be the asymptote for small x.

vela
Staff Emeritus
Homework Helper
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)?

Chestermiller
mfb
Mentor
That would be the asymptote for small x.
You don't need the linear term for this asymptote.

Chestermiller
Mentor
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).

mfb
Mentor
What you want to study is not a straight line. Let OP start with a simpler problem before we go to more advanced steps.

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.

mfb
Mentor
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.

Chestermiller
Mentor
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.

mfb
Mentor
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.

Chestermiller
Mentor
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.

mfb
Mentor
That is not the x-range where you would use an asymptote. It is between small x and large x.

Chestermiller
Mentor
That is not the x-range where you would use an asymptote. It is between small x and large x.
Whatever

SammyS
Staff Emeritus