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.

ok, this is how i get x=2.

x^2 = 2^x
2logx = xlog2
2/x. logx = log2
log(2/x)+ x = 2;
log(2/x) = 2-x
log x - log 2 = x-2;
logx - x = log 2 - 2;
so, x=2.

couldn't solve for x=4;

I got A in Calculus I, tara yo dekhe pachi ta malai feri precalculus padhnu parne jasto lagyo hehehehe....

good job guys. keep it up.
this kind of discussion is good for everyone.

.
x^2 = 2^x
2 (ln x) = x (ln 2)
(ln x)/x = (ln 2)/2

The following is the plot of the last expression, and you get two roots, 2 and 4.

It is interesting to see what happens if it is not x-squared but higher orders. It seems like there are only two real roots for everything more than (raised to the power) 2 except e itself, in which case we have only one real root. But there has to be more roots for higher order, right? Like cubic expressions should have 3 roots. I guess the ones we don't see are imaginary roots? As q tends to infinity, (ln q)/q tends to 0. Also, this line (ln q)/q first increases then decreases. Of course, for q<1, we have only one root - see the plot.

^ (ln q)/q increases then decreases with q for q>1

guys,
thanx a lot
for information, this equation doesn't just have the two positive roots..it does have a negative root also.like whatmore said, the higher degree eqns have more roots and the highest power of this equation is the variable itself.
so the solution is x=2,4 and -0.767
plz see the graph attached.........
havent taken the calc. 2 n 3
i believe log is the ultimate way for the solution...

looking at all these i dont' think i ever went to skool at all... Or maybe i went to skool to look at chicks..lol

but still how do you solve it algebraicly

I couldnt find the exact solution, not even with the help of mathematica.
That mathematica gave me one of the most complicated solution. Anyway
I tried to slove it in analytical way for the general case,

x^a=a^x ........(lets consider a is an integer and |a|>0)
log(a)(x^a)=log(a)(a^x)..(this is not natural logarihtm it is the logarithm with base a)
alog(a)(x)=xlog(a)(a)
alog(a)(x)=x.........(cause log(a)(a)=1)
a=x/log(a)(x).......................(1)

From equation 1 we can conclude following things.
1.x is an integer.
2.value of x follows the following progression(a,a^a,a^(a^a) and so on)
So it can be concluded from here that this equation can have
two real roots only if a^(a^(a^.........)) upto nth term/a^(a^(a^..........)) upto n-1th term=a
otherwise there is only one real root which is equal to a.

Hehe hope someone can come with better solution.

div
can u tell me what the mathmatica give u?

sujanks,
how did you get this??????????

log(2/x)+ x = 2;
log(2/x) = 2-x
log x - log 2 = x-2;

How about this??


x^2-2^x=0
x^2=2^x
log (x^2) = log (2^x)
2 log x = x log 2
log x / x = log 2 / 2
log x ^(1/x) = log 2 ^(1/2)

This is true only if x = 2

i guess i went a step ahead

log(2/x)+ x = 2;
log(2/x) = 2-x
->log2 - logx = 2-x
->logx - log2 - x-2

read fourth line as:
->logx - log2 = x-2

i mean how does 2/x. logx = log2 imply log(2/x)+ x = 2

can't.. sorry my solution is wrong... but i am sure there is a way to find it algebraicly... trying every possibility

well i m trying it myself from few days............
i told one of my frens to ask her prof in MO.........she asked 3 teachers but they couldn't do....now i m sending it to my high school teacher in LA (not Los Angeles.)

sujanks bro
thanks for your interest

Dunno how to prove it but try the following number all satisfy the criteria, short of time


x^2 - 2^x = 0

x= - 0.767 x = 1 x= 2 x= 4

Its been ages I got into these stupid puzzles, and dont waste time thinking. Just answers dears, dont have time. Appreciate it

I see all the graphs fail for x=1 bros so forget those stupid graphs

never mind bro
when x=1,
1^2 -2^1 =1-2
=-1
therefore x=1 is not the solution
the graph doesnt tell a lie