2006-07-30 02:40:08 · 18 answers · asked by Anonymous in Science & Mathematics ➔ Mathematics
x=2
x+x=2+2=4
x(x)=2(2)=4
x^x=2^2=4
2+2=2(2)=2^2
4=4=4
2 is called amicable number
2006-07-30 02:49:57 · answer #1 · answered by Yo! Mathematics 2 · 4⤊ 0⤋
X=1
2006-07-30 09:45:23 · answer #2 · answered by Albannach 6 · 0⤊ 0⤋
x=2
2006-07-30 19:22:33 · answer #3 · answered by talk2me_sister8 1 · 0⤊ 0⤋
x=2
2006-07-30 11:19:47 · answer #4 · answered by Mein Hoon Na 7 · 0⤊ 0⤋
x = 2
2006-07-30 09:44:09 · answer #5 · answered by Anonymous · 0⤊ 0⤋
x = 2
4 = 4 = 4
2006-07-30 09:43:46 · answer #6 · answered by Pyp 3 · 0⤊ 0⤋
x = 2?
2006-07-30 09:43:34 · answer #7 · answered by Ly L 2 · 0⤊ 0⤋
x is equal to 2
2+2=2 x 2=2^2=4
2006-07-30 11:43:48 · answer #8 · answered by the great man of lake mauvia 2 · 0⤊ 0⤋
2x=x^2=x^x.
2x=x^2
x=2.(if x is non-zero.)(x=0 is also a solution)
x^2=x^x.
As the bases are same, exponents are equal.
so, x=2.(Here x=0 is not a solution, because 0^2=0 but 0^0=1. We know x^0=1 for any x.)
x=2 is the only solution.
(I am not sure that 0^0 is 0 or 1. It should be 1 or not defined. I vote for 'not defined'. Because, I remember, (x^m)/(x^m)=1. (x^m)/(x^m)=x^(m-m)=x^0. Therefore x^0=1. This procedure does not hold good for x=0 as 0^m is 0 for a non-zero m. (x^m)/(x^m) is not defined for x=0. Hence 0^0 is not defined.)
2006-07-30 10:06:24 · answer #9 · answered by K N Swamy 3 · 1⤊ 0⤋
x=2
2+2=4
2*2=4
2^2=4
2006-07-30 09:44:13 · answer #10 · answered by Anonymous · 0⤊ 0⤋
x = 2
"Croasis" - Anything to the power 0 is 1. So 0^0 = 1
My mistake, the limit of x^0, as x tends to 0 is 1.
0^0 is undefined so x = 0 is not a solution.
I've just put 0^0 into Maple and it gave an answer of 1
I put 0^0 in Google and that gave 1
2006-07-30 10:12:15 · answer #11 · answered by Anonymous · 0⤊ 0⤋