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.

Hey Sujank,

You are looking for an algeraic solution, which is not possible for such type of equaiton. You have to solve them numerically as done by some of them above. The right answer is 2 and 4.
To solve it numerically you plot x (say from 0.1 to 5 in steps of 0.1) and y= logx/x-log2/2. And you find the place where the curve has a zero value, since y=0 gives the equation you need to solve.
So dont bang your head to find an exact numerical solution.

Sorry, I meant an exact algebraic expression in last line.

Ok little of matlab helped. I stll havent solved the problem.
but apparantly Lambart W function helps to solve this problem.
http://mathworld.wolfram.com/LambertW-Function.html

Also wikipedia gives one example
http://en.wikipedia.org/wiki/Lambert's_W_function

I got the following equation
-1/2log2=xlog(1/root(2))exp(xlog(root(2))
or x/2log2=W(-log2/2)
Apparantly W(-log2/2)=-log2
but W is supposed to be multivalued, now i didin get that
but using W(-log2/2)=-log2 we can at least get x=2

Well if someone can crack this whole thing it will be cool.

this is a very very intriguing problem

@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.

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

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.

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

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

I think I need my eyes checked. I read original poster posting," Can anyone solve: x^2 - 2^x =0 by algebraic method??? " :-s

Yea you indeed need to check your eyes,
Thats why you should do it algebraically and not by trail and error method!

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 ;-)

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.........

any more comments?

@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.

` 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.

the graphs only show the real solutions.............

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.

` you're right tregor. i still wonder why is it that the negative soln. is lost when we take logs.

You cannot take log of negative number but
log of a value can have negative number