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tregor
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Posted on 03-01-06 1:15
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can anyone solve: x^2 - 2^x =0 by algebraic method??? (x squared minus two to the power x equals zero) of course, without using graphing calculator.
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what more
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Posted on 03-03-06 11:28
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this is a very very intriguing problem
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divdude
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Posted on 03-04-06 12:32
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@crisna hehe see for example lets consider a general case 2^a-a^2=0 now here 'a' could be complex or a could be so huge that your computer cannot handle the calculation then you need analytical method. Fortunately the analytical method seems to exist . See my previous post.
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tregor
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Posted on 03-04-06 10:46
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Lambert's W function........yes absolutely.......i am used to the term "product log" but ... it's the same thing. i have seen people using product log to solve these kinda problems....... I have just finished my calc 1 so m dunno much abt using product log......but guess someone who have used it can solve the equation.....this certainly is not in my course but it's my curosity to get the way.... one of my cousins has half completed it by product log (Lambert's function) divdude,you are likely to make it
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crishna
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Posted on 03-04-06 11:19
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Hey dude, Solving in tems of a Lamberts function is not an analytical solution. If you get the answer in terms of a Lamberts function, you then have to solve the Lamberts function numerically agian. Let me explain why you cannot get an absolute analytical solution of that equaiton. To solve that equation one should be able to collect x in a place. You just cannot do it if you have log(x) or e^x or 2^x and x together i none equation. Why? because for example log(1+x)=x-x^2/2+x^3/3........and similar expansions for e^x and 2^x. It is impossible to extract x analytically to solve it analytically. You have to use some function like Lamberts or solve it numerically. Hope I made my point clear. Implementation of this in Matlab is easy.
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ChrisCornell
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Posted on 03-04-06 11:53
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kasto garo... x^2 - 2^x = 0 or, x^2 = 2^x or lgx^2 = lg2^x or 2lgx = xlg2 or x/lgx = 2/lg2 therefore, x = 2; however, x/lgx = 2/lg2 can also be expressed as: (2*2)/(2*lg2) (multiplying by 2/2=1) =4/(lg2^2) = 4/lg4 Giving x = 4. Result: x = (2,4) khoi k ho k ho....hisab bhanney pachi aafu lai pisab aaucha
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crishna
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Posted on 03-04-06 12:07
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Hey Chris, this is again not an analytical solution, what the original poster is looking for an analytical solution, that is, a way to solve algebraically, What you doing is putting x=2 and x=4, to make the equaiotn satisfied. If you dont know the answer(for eample in a complex version of the equation, your method of trial and guess wont work. I still insist every body not to look for abstract analyticcal solution, its just not possible. The best way to solve it is numerically or graphically
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ChrisCornell
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Posted on 03-04-06 12:15
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I think I need my eyes checked. I read original poster posting," Can anyone solve: x^2 - 2^x =0 by algebraic method??? " :-s
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crishna
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Posted on 03-04-06 12:25
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Yea you indeed need to check your eyes, Thats why you should do it algebraically and not by trail and error method!
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ChrisCornell
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Posted on 03-04-06 12:42
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J method bhaye ni baal ho; answer aaye pugyo. :D khoi class 8 padha algebra bhannya x , y patta lagauney bhanthiye...advance bhayecha ajkal...k k method k k method, kehi bujhey ta mori jaam ;-)
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tregor
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Posted on 03-04-06 3:30
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thanks chirscornell eeveryone was able to get the step logx/x=log2/2..........and conlude x=2 which i was not looking for but multiplying by 2/2 to get x=4 was good ....its kinda hit n trial but still appreciable.........
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tregor
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Posted on 03-06-06 1:24
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divdude
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Posted on 03-06-06 2:28
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@crisna I dont know much about lamebrt W function but it does have a proper algebrical expression. So getting solution in terms of Lamberts W function is equivalent to getting results in algebric form. For instance it is okay if we get result in log but to calculate actual log value you sitll have to get the value numerically. So I cant see why cant the solution of equation be expressed in Lamebrt's w function.
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what more
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Posted on 03-06-06 8:59
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` even if we take this Lambert function, we have to realize something - we lose information when we take logs. possible negative (and thus imaginary and complex) solutions are supressed as soon as the original equation is expressed in terms of logs. see the above graphs in previous posts.
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tregor
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Posted on 03-06-06 5:13
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the graphs only show the real solutions.............
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divdude
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Posted on 03-06-06 8:24
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I did a little study on Lambert w function. Lambert w function is actually a complex function and can also be expressed in form of laurant series. And it is multivalued i.e there are multiple solutions in complex plane but only 3 solutions in real line. About log i am not exactly sure but to say the least i remember this equation from Isc 2nd year, logx=x+x^2/1!+x^3/2!+......... so basically value of log can be found with good degree of precision. And you also can take log for complex numbers. But log being 200year old tradition, I dont really think it has too many flaws.
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what more
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Posted on 03-07-06 10:39
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` you're right tregor. i still wonder why is it that the negative soln. is lost when we take logs.
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divdude
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Posted on 03-07-06 11:04
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You cannot take log of negative number but log of a value can have negative number
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divdude
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Posted on 03-07-06 11:12
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Ah damm i forgot one small relation that i studied in complex analysis. Damm my memory. ok log(-1)=plus or minus i*PI
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tregor
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Posted on 03-07-06 2:58
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sujanks
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Posted on 03-07-06 4:13
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i don't think you can have log(-x), because this is base 10. for x to be -tive, the base must be and has to be an odd number.
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