2007-03-16 03:05:52 · 6 answers · asked by Minerva 1 in Science & Mathematics ➔ Mathematics
2^x+2^(-x)=3
(2^x+2^(-x))^2=3^2
2^2x + 2*2^x*2(-x) + 2^(-2x) =9
4^x+4^(-x)+2=9
4^x+4^(-x)=7
2007-03-16 03:15:58 · answer #1 · answered by diamond 3 · 1⤊ 1⤋
Try multiplying out (2^x + 2^(-x))^2 = 9. You will find that this leads easily to 4^x + 4^(-x) = an integer. (Hint 4 = 2^2)
2007-03-16 10:14:05 · answer #2 · answered by mathsmanretired 7 · 1⤊ 1⤋
2^x + 2^(-x) = 3
square both sides:
(2^x)^2 + 2 * 2^x * 2^-x + (2^-x)^2 = 9
(2^x)^2 + 2 * 2^(x - x) + (2^-x)^2 = 9
(2^x)^2 + 2 * 2^0 + (2^-x)^2 = 9
(2^x)^2 + 2 * 1 + (2^-x)^2 = 9
(2^x)^2 + 2 + (2^-x)^2 = 9
(2^x)^2 + (2^-x)^2 = 9 - 2
(2^x)^2 + (2^-x)^2 = 7
2^2x + 2^-2x = 7
(2^2)^x + (2^2)^(-x) = 7
4^x + 4^(-x) = 7
2007-03-16 10:16:29 · answer #3 · answered by Mathematica 7 · 0⤊ 0⤋
7
4^x+4^(-x) = 2^2x+2^(-2x)
Because 2^2 = 4
so sqr the first equation and simply:
(2^x+2^(-x))^2=(3)^2
2^2x + 2 + 2^-2x = 9
2^2x + 2^-2x = 7
since 2^2x + 2^-2x = 4^x+4^(-x)
your answer is 7
2007-03-16 10:21:18 · answer #4 · answered by jpferrierjr 4 · 0⤊ 0⤋
try to square the whole equation and you will get 4^x+4^(-x)=7,since 2^(2x)=4^x
2007-03-16 10:25:38 · answer #5 · answered by Kelvin C 1 · 0⤊ 0⤋
I see everyone has the right answer now. So, who was the first one to post an answer of 7 rather than 9? Be honest....
2007-03-16 10:36:43 · answer #6 · answered by Anonymous · 0⤊ 0⤋