Math problem

[MisterPants]

How do I solve x^(1/2) = e^(-3x)? I seem to be twisting it in circles...

 

[CubicZirconia]

Use a natural log.

 

[phatj]

If my logarithmic memory serves me right...
ln x^(1/2) = ln e ^ (-3x)..

ln x^(1/2) = -3x

umm... sorry im on summer vacation

 

[MisterPants]

All I can manage is:
1/2lnx=-3x
ln[x^(1/2)e^(3x)]=0
x^(1/2)e^3x=1
Which takes me back to the beginning...

 

[wyvrn]

The ln of e to any power = that power. So, ln e^-3x = -3x. If you take the ln of both sides, you end up with ln x^1/2 = -3x right? BTW, you cannot make ln x^1/2 = 1/2 ln x, because 1/2 is the power of x and not a coefficient, so what you have done is flawed. I am not sure what to do after ln x^1/2 = -3x though, I will think on it and get back.

Hm, then you could divide by -3 to get {-1/3 * ( ln x^1/2) } = x , not sure how to simplify beyond that.

 

[GoingUp]

do you have a graphic calculator?

graph x^1/2 and e^-3x and calculate the intersection

 

[CubicZirconia]

nm, this would be easier if I was still in calc.

 

[Indolent]

Originally posted by: wyvrn
The ln of e to any power = that power. So, ln e^-3x = -3x. If you take the ln of both sides, you end up with ln x^1/2 = -3x right? BTW, you cannot make ln x^1/2 = 1/2 ln x, because 1/2 is the power of x and not a coefficient, so what you have done is flawed. I am not sure what to do after ln x^1/2 = -3x though, I will think on it and get back.

Hm, then you could divide by -3 to get {-1/3 * ( ln x^1/2) } = x , not sure how to simplify beyond that.

taking the natural log moves the power of x to the coefficient....simple natural log rule

*edit* which also proves you're first statement. ln e^-3x = -3x ln(e) ....... ln(e)=1 so it cancels out, leaving -3x


*edit^2* my calculator says .238 but i'm not quite sure how to get to that answer

 

[Turkish]

argh math hates me and i hate math

 

[AznMaverick]

trial and error.

 

[Cycad]

All your answers here.

 

[GiLtY]

One is a transcedantal function while the other is an algebraic. The only thing I can think of is calculator, or my calculus is failing me...

 

[clamum]

Originally posted by: Cycad
All your answers here.

LOL

 

[BruinEd03]

Originally posted by: MisterPants
How do I solve x^(1/2) = e^(-3x)? I seem to be twisting it in circles...

try graphing it.

-Ed

 

[Special K]

I don't think you can solve it analytically:

Dr. Math question similar to yours

 

[Muzzan]

Originally posted by: wyvrn
The ln of e to any power = that power. So, ln e^-3x = -3x. If you take the ln of both sides, you end up with ln x^1/2 = -3x right? BTW, you cannot make ln x^1/2 = 1/2 ln x, because 1/2 is the power of x and not a coefficient, so what you have done is flawed. I am not sure what to do after ln x^1/2 = -3x though, I will think on it and get back.

Hm, then you could divide by -3 to get {-1/3 * ( ln x^1/2) } = x , not sure how to simplify beyond that.

Nope, ln(sqrt(x)) = ln(x)/2. Follows from the definition of the logarithm

ln(a^b) = b * ln(a). But if a happens to be e, then ln(a) is 1 (which is what you said in your first sentence, that the natural log of e to any power = that power).

 

[wyvrn]

Ok so the answer is then (-1/6 ln x) = x, but where do we go from here?

Originally posted by: Muzzan

Originally posted by: wyvrn
The ln of e to any power = that power. So, ln e^-3x = -3x. If you take the ln of both sides, you end up with ln x^1/2 = -3x right? BTW, you cannot make ln x^1/2 = 1/2 ln x, because 1/2 is the power of x and not a coefficient, so what you have done is flawed. I am not sure what to do after ln x^1/2 = -3x though, I will think on it and get back.

Hm, then you could divide by -3 to get {-1/3 * ( ln x^1/2) } = x , not sure how to simplify beyond that.

Nope, ln(sqrt(x)) = ln(x)/2. Follows from the definition of the logarithm

ln(a^b) = b * ln(a). But if a happens to be e, then ln(a) is 1 (which is what you said in your first sentence, that the natural log of e to any power = that power).

 

[Muzzan]

wyvrn: you can't go anywhere from there, since the equation can't be solved algebraically. You can graph it or use, say, Newton-Rapson to get an approximation though.

 

[phatj]

the natural log of e is 1. That is all i have left to contribute to this thread

 

[Kyteland]

Equivalent equation: x*e^6x = 1

Unfortunately this can only be approximated. You could try integrating x*e^6x-1 = 0 and solving forany local maximum/minimum values, but I honestly don't see how this method would give you anything other than another approximation.

 

[rgwalt]

First, graph the problem in order to see how many solutions there are. Then you can use Newton's method, or you can do a fixed point iteration, which is even easier.

Ryan

 

[Woodchuck2000]

Newton-Raphson is by far your best bet... Short of drawing the graph and counting squares

 

[dmw16]

If you have matlab it could solve it easily for you. Or if you knew some C++ you could write a brute force program to find the answer. Or, if I remember right there is something called a W function, but I dont really remember how it works. Good luck,
-doug

 

[wyvrn]

Ok above my head. Didn't learn Newton-Rapson in my two years of calc.

Originally posted by: Muzzan
wyvrn: you can't go anywhere from there, since the equation can't be solved algebraically. You can graph it or use, say, Newton-Rapson to get an approximation though.

 

[lowfatbaconboy]

ti-89