Solve for x --->> 2^(1+x) + 4^x = 4 - 2^x?

2^(1+x) + 4^x = 4 - 2^x

2^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

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

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

x=0

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

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^(1+x)+2^2x=2^2-2^x
1+x+2x=2+x
1=-2x
x=-1/2

Simple inspection reveals that x = 0

2^(1+x) + 2^2x = 2^2 - 2^x
(1+x) + 2 = 2 - x
x + x = 2 - 2 - 1
2x = -1
x = -1/2


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