factor out 2^x, after bringing 12 to the other side:
2^x(2^x+2^2) = 12. Let y = 2^x. Then y(y+4) = 12, or y^2+4y-12 = 0.
So (y+6)(y-2) = 0. y cannot be negative, however, according to its definition. Therefore, y = 2, 2^x = 2, x=1.
Steve
2007-01-28 19:14:24
·
answer #1
·
answered by Anonymous
·
1⤊
0⤋
2^2x + 2^(x+2) - 12 = 0
2^2x + 2^x(2^2) - 12 = 0
2^2x + 4*2^x - 12 = 0
(2^x + 6)(2^x - 2) = 0
2^x = -6, 2
log[base 2](2^x) = log[base2](2)
-6 is rejected as a solution since you can't take a log of a negative number.
x = 1
2007-01-28 19:43:55
·
answer #2
·
answered by Northstar 7
·
0⤊
0⤋
You can get everything in terms of 2^x using rules for manipulating exponents.
2^2x = (2^x)^2, and
2^(x+2) = 2^2 * 2^x
So you have a quadratic equation in 2^x. If you substitute u for 2^x, you get
u^2 + 4u - 12 = 0, which I'm sure you can solve for u. One of the results is negative, and 2^x cannot be negative if x is real, so set it equal to the positive result for u, and you'll get that answer for x ... as you said, 1.
2007-01-28 19:23:05
·
answer #3
·
answered by Hal 2
·
1⤊
0⤋
2^2x+2^(x+2)-12=0
=2^2x+2^x*2^2-12=0
= (2^x)^2+2^x*4-12=0
let 2^x=y
y^2+4y-12=0
y^2+6y-2y-12=0
y(y+6)-2(y+6)=0
(y-2)?(y+6)=0
so y=2 or y= -6
so 2^x=1 or -6
as 2^x can not be negativeo ignore -6
now 2^x=2
so x=1
2007-01-28 19:17:30
·
answer #4
·
answered by Pradeep Chelani 2
·
0⤊
0⤋
let 2^x =t
then
t^2+4t-12=0
t1=2
t2=-6
2^x=2^1
hence x=1
2^x= -6
does not satisfy
2007-01-28 19:17:11
·
answer #5
·
answered by iyiogrenci 6
·
0⤊
0⤋