Question

solve the equation

Answer 1

\[(2^y-y^2)^3=2^y-y^2\]

Answer 2

\[(2^y−y^2)^3=2^y−y^2\]\[\frac{(2^y−y^2)^3}{2^y−y^2}=\frac{2^y−y^2}{2^y−y^2}\]\[(2^y−y^2)^2=1\]\[2^{2y}-2\times2^yy^2+y^4=1\]\[2^{2y}-2^{y+1}y^2+y^4-1=0\]\[...\]

Answer 3

\[(2^y-y^2)\]may equal to zero

Answer 4

Can plot and solve?

Answer 5

no

Answer 6

\[ (2^y−y^2)^2 - 1 = 0 \]
\[ (2^y−y^2 - 1)(2^y−y^2 + 1) = 0 \]
Solving them individually would be better. I think

Answer 7

\( 2^y - y^2 + 1 = 0 \implies y = 3\)

Answer 8

The best way would be to use solve it Graphically ... or use Newton's method ... I think!!

Answer 9

Do you have another method other than Graphically and Newton's method?

Answer 10

try something like this
http://math.stackexchange.com/questions/127199/how-to-solve-an-exponential-equation-with-two-different-bases-3x-2x-5

Answer 11

Looks more like numeric method than algebraic method ... if you find another method ... i would also like to know!!

Answer 12

(2^y – y^2)^3 – (2^y – y^2) =0
(2^y – y^2) [(2^y – y^2)^2 – 1] =0
(2^y – y^2) (2^y – y^2 – 1) (2^y – y^2 + 1) =0
2^y – y^2 = 0 or 2^y – y^2 = 1 or 2^y – y^2 = –1

case 1,
2^y – y^2 = –1
2^y = y^2 – 1
2^y = (y + 1 )(y – 1)
the difference between (y+1) and (y-1) is 2
they must be both even or both odd.
the product of (y+1) and (y-1) is 2^y which is even.
so, (y+1) and (y-1) are both even and y must be odd.

the single-digit of 2^y is 2 or 4 or 8 or 6
Since 2^y = y^2 – 1, the single-digit of y^2 -1 is 2 or 4 or 8 or 6
the single-digit of y^2 must be 3 or 5 or 9 or 7
because of y^2 , the single-digit of y^2 must not be 3 and 7.
So, the single-digit of y^2 must be 5 or 9

if y is more , 2^y >> y^2
Thus y must equal to 3 or 5 or 7.

Therefore, y=3

Answer 13

looks like I completely ignored 2

Answer 14

moreover we could hardly call it algebraic .... i think it's more analytic

Answer 15

This method cannot find non-integer solution :(

Answer 16

also we are missing zero!!
http://www.wolframalpha.com/input/?i=%282^y+-+y^2%29^3+%3D+%282^y+-+y^2%29

Answer 17

Oo ... 0,1,2,3,4 are all integer solutions <-- wolf says so!!

Answer 18

2^y – y^2 = 1
2^y = y^2 + 1
the single-digit is 1 or 2or 6
y=0 or y=1

Answer 19

2^y – y^2 = 0 => y = 2, 4

Answer 20

it seems that power function is always less than exponentially increasing function ... so
2^y and y^2 intersects at at most two points!!
using it logically, we can find the solution of the latter as 2 and 4