# Trying to solve complex equation with 1 variable,

• QuantumX
In summary, the equation gives the radius of the dark matter halo for the Milky Way. It is supposed to give a value of 100 kpc, but the user is having trouble solving it. They ask for help from their peers, and one of them suggests using WolframAlpha. They plug in the given expression and get a result in kilometers.
QuantumX

## Homework Statement

$220000=\sqrt{1.4*10^{32}*\left[ln(\frac{6.17*10^{20}+x}{6.17*10^{20}})-\frac{x}{6.17*10^{20}+x}\right]*\frac{1}{x}}$

It's supposed to give me the radius of the Milky Way's dark matter halo. I expect to get a value of around 100 kpc (equation will produce a value in m), although that would assume I didn't mess up anywhere in deriving it.

Please don't ask me to show you how I derived it, I mean we can do that, but right now I'd really just like to get a value for x. I just can't solve it myself, and all online calculators I've tried it on have failed.

Can you please help me solve it, it took me all day to get here and now I can't get the actual value.

Thanks!

Show us what steps you've already taken. If you are looking for a totally algebraic solution, where you get x = some number, that doesn't appear to be in the cards. It'll probably take an iterative method to find x.

QuantumX said:

## Homework Statement

$220000=\sqrt{1.4*10^{32}*\left[ln(\frac{6.17*10^{20}+x}{6.17*10^{20}})-\frac{x}{6.17*10^{20}+x}\right]*\frac{1}{x}}$

It's supposed to give me the radius of the Milky Way's dark matter halo. I expect to get a value of around 100 kpc (equation will produce a value in m), although that would assume I didn't mess up anywhere in deriving it.

Please don't ask me to show you how I derived it, I mean we can do that, but right now I'd really just like to get a value for x. I just can't solve it myself, and all online calculators I've tried it on have failed.

Can you please help me solve it, it took me all day to get here and now I can't get the actual value.

Thanks!
Graph the function.

I can't, I don't have any sort of math software, even if I did I doubt it would do it haha.

It 's supposed to be a value, but I can see how it would be impossible to get from this.

Still, any help is appreciated.

QuantumX said:
I can't, I don't have any sort of math software, even if I did I doubt it would do it haha.

It 's supposed to be a value, but I can see how it would be impossible to get from this.

Still, any help is appreciated.
Try WolframAlpha .

Basically, the way I derived it (and this may now be more appropriate for the astrophysics section)

is by using the equation

$V=\sqrt{\frac{GM}{R}}$

I want to estimate the point at which orbital velocity and mass are no longer linearly related. The velocity of the flat portion of the Milky Way's rotation curve is around 220,000m/s. I am looking for a radius Rmax at which the orbital velocity drops off, and that would be the edge of the dark matter halo.

So I have V, I'm looking for R and I now need M to substitute in the above equation.

$M=4/3\pi r^3ρ(r)$

Integrating:

$M=∫ 4 \pi r^2ρ(r)dr = 4\pi ρ_{0}Rs^3[ln(\frac{Rs + Rmax}{Rs}) - \frac{Rmax}{Rs+Rmax}]$

I have values for ρ(0) and Rs (those are constants for our galaxy)

So substituting everything, I get the equation you see.

@SammyS - Thank you so much, that worked. I got a graph, but I'm not sure how to interpret it...

Here's the equation in a form that works for WolframAlpha

220000=sqrt(1.4*10^32*(ln((6.17*10^20+x)/(6.17*10^20))- (x/(6.17*10^20+x))*(1/x))

Yeah, I guess it's a function because the radius is a function of the density, but I thought for some reason that problem would be solved by the fact I had the density constants for our galaxy. I wish WolframAlpha would show me more of the graph, so I can see where it gets near flat, that's what I'll call x for my estimate.

I just totally have no idea what I'm doing haha, but it's been fun.

QuantumX said:
@SammyS - Thank you so much, that worked. I got a graph, but I'm not sure how to interpret it...

Here's the equation in a form that works for WolframAlpha

220000=sqrt(1.4*10^32*(ln((6.17*10^20+x)/(6.17*10^20))- (x/(6.17*10^20+x))*(1/x))

Yeah, I guess it's a function because the radius is a function of the density, but I thought for some reason that problem would be solved by the fact I had the density constants for our galaxy. I wish WolframAlpha would show me more of the graph, so I can see where it gets near flat, that's what I'll call x for my estimate.

I just totally have no idea what I'm doing haha, but it's been fun.
Well, you could change x to units of pc or kpc. (I assume x is in meters.)

Or try this for the given expression. (I divided both sides or your equation by 220000 then subtracted 1.) https://www.wolframalpha.com/input/?i=plot+%281.18322%2F220000%29*10%5E16+sqrt%28%28-x%2F%286.17x10%5E20%2Bx%29%2Blog%281.62075x10%5E-21+%286.17x10%5E20%2Bx%29%29%29%2Fx%29-1+for+x+from+8*10%5E20+to+2+*+10%5E21

QuantumX said:

## Homework Statement

$220000=\sqrt{1.4*10^{32}*\left[ln(\frac{6.17*10^{20}+x}{6.17*10^{20}})-\frac{x}{6.17*10^{20}+x}\right]*\frac{1}{x}}$

It's supposed to give me the radius of the Milky Way's dark matter halo. I expect to get a value of around 100 kpc (equation will produce a value in m), although that would assume I didn't mess up anywhere in deriving it.

Please don't ask me to show you how I derived it, I mean we can do that, but right now I'd really just like to get a value for x. I just can't solve it myself, and all online calculators I've tried it on have failed.

Can you please help me solve it, it took me all day to get here and now I can't get the actual value.

Thanks!
Let y = x /6.17 x 1020
Then:
$220000=\sqrt{\frac{2.27\times 10^{11}}{y}*\left[ln(1+y)-\frac{y}{1+y}\right]}$
Square both sides:
$$0.213=\left[\frac{ln(1+y)}{y}-\frac{1}{1+y}\right]$$
The solution to this equation is approximately y = 1.5.

Chet

Last edited:
Hm, well I'm not sure how to interpret yours.

When I look at mine:

https://www.wolframalpha.com/input/...0+x)/(6.17*10^20))-+(x/(6.17*10^20+x))*(1/x))

I see that the curve for x ends right at 3.75*10^21 m = 121.5 kpc, which is within the range of current estimates for the radius of the galaxy (the dark matter halo that is).

That's just an observation, I' not sure how to read the graph honestly. Does the curve end where the graph ends, or does it continue?

I know that x is radius, but what is y ? Is it density? That wouldn't make sense, since density is in kg/m^3, which would be way too much according to the graph for that radius.

Sorry for the super newbie questions.

Chestermiller said:
Let y = x /6.17 x 1020
Then:
$220000=\sqrt{\frac{2.27\times 10^{11}}{y}*\left[ln(1+y)-\frac{y}{1+y}\right]}$
Square both sides:
$$0.213=\left[\frac{ln(1+y)}{y}-\frac{1}{1+y}\right]$$
The solution to this equation is approximately y = 1.5.

Chet

Hm, that only gives 30 kpc for the radius, it's supposed to be around 10 times that.

QuantumX said:
Hm, that only gives 30 kpc for the radius, it's supposed to be around 10 times that.
Sorry about that. Substitute the result, and see if it satisfies your equation.

Chet

That's what I did, it's off by a factor of 10, but I'm sure the error is in my calculations somewhere.

Chestermiller said:
Let y = x /6.17 x 1020
Then:
$220000=\sqrt{\frac{2.27\times 10^{11}}{y}*\left[ln(1+y)-\frac{y}{1+y}\right]}$
Square both sides:
$$0.213=\left[\frac{ln(1+y)}{y}-\frac{1}{1+y}\right]$$
The solution to this equation is approximately y = 1.5.

Chet
Here's the WA graph for ##\displaystyle \left[\frac{ln(1+y)}{y}-\frac{1}{1+y}\right]- 0.213 ##

So, there are two solutions.

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## 1. How do you approach solving a complex equation with 1 variable?

The first step in solving a complex equation with 1 variable is to simplify the equation by combining like terms and using the order of operations. This will make the equation easier to work with.

## 2. What is the best method for solving a complex equation with 1 variable?

The method for solving a complex equation with 1 variable will depend on the specific equation. Some common methods include substitution, elimination, or graphing. It is important to choose a method that will simplify the equation and make it easier to solve.

## 3. How do you check if your solution to a complex equation with 1 variable is correct?

To check if your solution is correct, you can substitute the value you found for the variable back into the original equation. If the equation is true, then your solution is correct. You can also use a graphing calculator to graph the equation and see if the solution falls on the curve.

## 4. Can a complex equation with 1 variable have more than one solution?

Yes, a complex equation with 1 variable can have more than one solution. This is known as having multiple roots. It is important to carefully check your work and consider all possible solutions when solving an equation.

## 5. Are there any tips for solving complex equations with 1 variable more efficiently?

One tip for solving complex equations with 1 variable more efficiently is to practice regularly and become familiar with different methods for solving equations. It can also be helpful to break the equation down into smaller, simpler steps and to check your work as you go along. Additionally, using a graphing calculator or software can speed up the process of solving equations.

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