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Math problem

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
 
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.
 
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.
Click to expand...

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
 
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...
 
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.
Click to expand...

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).
 
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.
Click to expand...

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).
Click to expand...

 
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.
 
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.
 
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
 
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
 
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.
Click to expand...

 
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