Please Solve for X:
2^2x - 2^(x+1) - 8 = 0
2007-07-28 21:37:23 · 5 answers · asked by im_n0t 1 in Science & Mathematics ➔ Mathematics
assume 2^x as another variable let say 'y''
now the equation becomes
y^2-2y-8=0
which is
y-4 )* y+2) =0
which gives y=4 and y=-2
now y=2^x
which gives 2^x=4 and 2^x=-2
but 2^x=-2 doesnt hav any solution
thus the only solution is
2^x=4 >>> x=2
2007-07-28 21:44:57 · answer #1 · answered by titu-the matrix 2 · 0⤊ 0⤋
Let y = 2^x
then 2^2x - 2^(x+1) - 8 = 0
becomes
y^2 - 2y - 8 = 0
(y - 4) (y + 2) = 0
y = 4 or y = -2
2^x = 4 ===>>> x = 2
2^x = -2 has no solution,
so x=2 is the only answer
2007-07-29 07:15:13 · answer #2 · answered by vlee1225 6 · 0⤊ 0⤋
Let y = 2^x
y² - 2 y - 8 = 0
(y - 4 ) (y + 2) = 0
y = 4 , y = - 2
Accept y = 4
2^x = 4
x = 2
2007-07-29 08:12:59 · answer #3 · answered by Como 7 · 0⤊ 0⤋
Let 2^x = y
The eqn becomes
y^2 - 2y - 8 = 0
(y+2)(y-4) = 0
y = -2 is not possible
Thus y = 4 or 2^x = 4 or x = 2
2007-07-29 04:41:42 · answer #4 · answered by ag_iitkgp 7 · 0⤊ 2⤋
2x^2 - (x+1)^2 - 8 = 0
2x^2 - (x^2+2x+1) - 8 = 0
2x^2 -x^2 - 2x -1 - 8 = 0
x^2 - 2x = 9
x^2 - 2x + (1)^2 = 9 + 1
(x - 1)^2 = 10
sqrt [(x - 1)^2] = sqrt (10)
x - 1 = +-sqrt(10)
x = sqrt(10) + 1
or
x = 1 - sqrt(10)
2007-07-29 04:47:30 · answer #5 · answered by fofo m 3 · 0⤊ 0⤋