it's rare ...tough question.....i think it 's impossible
2007-03-04 21:32:50 · 11 answers · asked by adeeba j 1 in Science & Mathematics ➔ Mathematics
log2(2^x) = 1+log2(x)
x-1=log2(x)
now expan log2x and solve
2007-03-04 21:40:54 · answer #1 · answered by Anonymous · 0⤊ 0⤋
x=2
2007-03-05 05:41:10 · answer #2 · answered by Sapph 3 · 0⤊ 1⤋
x=1
2007-03-05 05:36:19 · answer #3 · answered by HeAvEnLy_PiNk 3 · 0⤊ 1⤋
Yes x can be both 1 and 2
2007-03-05 05:59:25 · answer #4 · answered by Ali 3 · 1⤊ 0⤋
it can be both 1 or 2
2007-03-05 05:47:48 · answer #5 · answered by wizkid!!! 3 · 1⤊ 0⤋
By inspection the values x = 1 and x = 2 satisfy the equation.
2007-03-05 06:13:14 · answer #6 · answered by physicist 4 · 1⤊ 0⤋
There are only two values of x whixh satisfy this equation, namely x = 1 and x = 2. It does not therefore look as impossible as it appears to you.
2007-03-05 06:47:10 · answer #7 · answered by Paleologus 3 · 0⤊ 0⤋
x can either be 1 or 2
2007-03-05 06:06:24 · answer #8 · answered by Twarita 2 · 1⤊ 0⤋
2
2 squared = 2 x 2 = 4
we do the impossible, quickly, here
and we are humble
2007-03-05 05:35:58 · answer #9 · answered by tomkat1528 5 · 0⤊ 1⤋
4^x
2007-03-05 05:37:04 · answer #10 · answered by richard_lee_boyd 2 · 0⤊ 1⤋