Solve for x --->> 2^(1+x) + 4^x = 4 - 2^x?
回答 (9)
2009-02-18 2:17 pm
✔ 最佳答案
2^(1+x) + 4^x = 4 - 2^x2^x = y
2^(1+x) = 2y
4^x = y²
2y + y² = 4 - y
y² + 3y - 4 = 0
(y+4)(y-1) = 0
y = {-4, 1}
But y>0
y = 1
2^x = 1
x = 0
2016-10-18 9:40 pm
first multply each thing by making use of 4x^2 - forty 9: (4x^2 - forty 9 / 2x - 7) + (x - 5) = 4(4x^2 - forty 9) / (6x - 21) (2x - 7)(2x + 7) / (2x - 7) + (x - 5) = 4(2x - 7)(2x + 7) / 3(2x - 7) 2x + 7 + x - 5 = 4(2x + 7) / 3 3x + 2 = 4(2x + 7) / 3 9x + 6 = 8x + 28 x = 22
2009-02-18 2:57 pm
2^(1 + x) + 4^x = 4 - 2^x
2^(1 + x) + (2^2)^x = 2^2 - 2^x
2^(1 + x) + 2^(2x) = 2^2 - 2^x
2^(x + 1) + 2^(2x) + 2^x = 2^2
Let 2^x = y.
2^(x + 1) + 2^(2x) + 2^x = 2^2
2^x(2^1) + (2^x)(2^x) + 2^x = 2^2
y(2^1) + (y)(y) + y = 4
2y + y^2 + y - 4 = 0
y^2 + 3y - 4 = 0
y^2 + 4y - y - 4 = 0
(y^2 + 4y) - (y + 4) = 0
y(y + 4) - 1(y + 4) = 0
(y + 4)(y - 1) = 0
y + 4 = 0
y = -4
y - 1 = 0
y = 1
Change y into 2^x.
y = -4
2^x = -4
x = log_2(-4)
(undefined)
y = 1
2^x = 1
2^x = 2^0
x = 0
∴ x = 0
2^(1 + x) + (2^2)^x = 2^2 - 2^x
2^(1 + x) + 2^(2x) = 2^2 - 2^x
2^(x + 1) + 2^(2x) + 2^x = 2^2
Let 2^x = y.
2^(x + 1) + 2^(2x) + 2^x = 2^2
2^x(2^1) + (2^x)(2^x) + 2^x = 2^2
y(2^1) + (y)(y) + y = 4
2y + y^2 + y - 4 = 0
y^2 + 3y - 4 = 0
y^2 + 4y - y - 4 = 0
(y^2 + 4y) - (y + 4) = 0
y(y + 4) - 1(y + 4) = 0
(y + 4)(y - 1) = 0
y + 4 = 0
y = -4
y - 1 = 0
y = 1
Change y into 2^x.
y = -4
2^x = -4
x = log_2(-4)
(undefined)
y = 1
2^x = 1
2^x = 2^0
x = 0
∴ x = 0
2009-02-18 2:20 pm
x=0
2009-02-18 2:20 pm
2 * 2^x + 2^x * 2^x = 4 - 2^x <=>
if X = 2^x then 2 * X + X² = 4 - X <=> X² + 3 *X - 4 = 0
Delta = 9 + 16 = 25
X1 = (-3 - 5) / 2 = -4 and X2 = (-3 + 5) / 2 = 1
Only X2 is solution of the above equation because 2^x > 0
X2 = 1 <=> 2^x = 1 <=> x ln 2 = ln 1 <=> x = ln 1 / ln 2 = 0
Solution : x = 0
if X = 2^x then 2 * X + X² = 4 - X <=> X² + 3 *X - 4 = 0
Delta = 9 + 16 = 25
X1 = (-3 - 5) / 2 = -4 and X2 = (-3 + 5) / 2 = 1
Only X2 is solution of the above equation because 2^x > 0
X2 = 1 <=> 2^x = 1 <=> x ln 2 = ln 1 <=> x = ln 1 / ln 2 = 0
Solution : x = 0
2009-02-18 2:18 pm
2^1.2^x + 2^(2x) = 4 - 2^x
2.2^x + 2^(2x) = 4 - 2^x
2^(2x) + 2.2^x + 2^x - 4 = 0
2^(2x) + 3.2^x - 4 = 0
Let y = 2^x, then y^2 = 2^(2x)
y^2 + 3y - 4 = 0
(y - 1) (y + 4) = 0
y = 1 or y = -4
2^x = 1
2^x = 2^0
x = 0
2^x =/= -4
Solution set = {0}.
2.2^x + 2^(2x) = 4 - 2^x
2^(2x) + 2.2^x + 2^x - 4 = 0
2^(2x) + 3.2^x - 4 = 0
Let y = 2^x, then y^2 = 2^(2x)
y^2 + 3y - 4 = 0
(y - 1) (y + 4) = 0
y = 1 or y = -4
2^x = 1
2^x = 2^0
x = 0
2^x =/= -4
Solution set = {0}.
參考: Dont Ask...
2009-02-18 2:14 pm
Simple inspection reveals that x = 0
2009-02-18 2:20 pm
2^(1+x) + 2^2x = 2^2 - 2^x
(1+x) + 2 = 2 - x
x + x = 2 - 2 - 1
2x = -1
x = -1/2
anvil ltd.
(1+x) + 2 = 2 - x
x + x = 2 - 2 - 1
2x = -1
x = -1/2
anvil ltd.
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