Log-Plots: Solving Homework Equations

[WWCY]

Homework Statement

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!

 

[mfb]

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]

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.

 

[WWCY]

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]

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.

 

[WWCY]

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.

 

[Chestermiller]

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]

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)?

 

[WWCY]

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]

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

 

[mfb]

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]

That would be the asymptote for small x.

 

[vela]

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)?

 

[mfb]

That would be the asymptote for small x.

You don't need the linear term for this asymptote.

 

[Chestermiller]

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]

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

 

[WWCY]

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]

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]

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]

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]

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]

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

 

[Chestermiller]

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

Whatever

 

[SammyS]

View attachment 209498

That looks like asymptotic behavior to me !