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 solve this equation

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Posted on 03-01-06 1:15 PM     Reply [Subscribe]
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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.
 
Posted on 03-01-06 8:32 PM     Reply [Subscribe]
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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;
 
Posted on 03-01-06 8:56 PM     Reply [Subscribe]
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I got A in Calculus I, tara yo dekhe pachi ta malai feri precalculus padhnu parne jasto lagyo hehehehe....
 
Posted on 03-01-06 9:22 PM     Reply [Subscribe]
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good job guys. keep it up.
this kind of discussion is good for everyone.
 
Posted on 03-02-06 12:37 AM     Reply [Subscribe]
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.
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.

 
Posted on 03-02-06 12:45 AM     Reply [Subscribe]
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^ (ln q)/q increases then decreases with q for q>1
 
Posted on 03-02-06 11:55 AM     Reply [Subscribe]
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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...

 
Posted on 03-02-06 12:55 PM     Reply [Subscribe]
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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
 
Posted on 03-02-06 2:57 PM     Reply [Subscribe]
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but still how do you solve it algebraicly
 
Posted on 03-02-06 9:11 PM     Reply [Subscribe]
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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.
 
Posted on 03-03-06 12:24 PM     Reply [Subscribe]
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div
can u tell me what the mathmatica give u?
 
Posted on 03-03-06 12:31 PM     Reply [Subscribe]
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sujanks,
how did you get this??????????

log(2/x)+ x = 2;
log(2/x) = 2-x
log x - log 2 = x-2;
 
Posted on 03-03-06 12:39 PM     Reply [Subscribe]
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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
 
Posted on 03-03-06 1:31 PM     Reply [Subscribe]
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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
 
Posted on 03-03-06 1:32 PM     Reply [Subscribe]
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read fourth line as:
->logx - log2 = x-2
 
Posted on 03-03-06 1:50 PM     Reply [Subscribe]
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i mean how does 2/x. logx = log2 imply log(2/x)+ x = 2
 
Posted on 03-03-06 2:25 PM     Reply [Subscribe]
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can't.. sorry my solution is wrong... but i am sure there is a way to find it algebraicly... trying every possibility
 
Posted on 03-03-06 4:25 PM     Reply [Subscribe]
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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.)
 
Posted on 03-03-06 4:26 PM     Reply [Subscribe]
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sujanks bro
thanks for your interest
 
Posted on 03-03-06 4:41 PM     Reply [Subscribe]
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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
 
Posted on 03-03-06 4:59 PM     Reply [Subscribe]
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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
 



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