Como resolve a equação: 4^x+1-33*2^x+8=0 ?.... Pergunta de ideia deLeticiaamaro12

ellencristinad 4^(x+1) - 33.2^x + 8 = 0
4.4^x - 33.2^x +8 = 0........=> 4^x=y²...e....2^x=y
4y² -33y +8 = 0
D=b² -4.a.c = (-33)² -4.4.8 =1089 -128= 961....V961=31
y'=-b+VD/2a = 33+31/2.4= 64/8 = 8
y''=-b-VD/2a = 33-31/2.4= 2/8 = 1/4

2^x' = y'
2^x'= 8
2^x = 2^3.......=> x'=3 #

2^x''=y''
2^x'' = 1/4
2^x''= 2^-2.......=> x''= -2 #

S: { -2, 3 }

10 votes Thanks 15

ellencristinad 4^(x+1) - 33.2^x + 8 = 0 2^(2x+2) - 33.2^x + 8 = 0 4*2^2x - 33.2^x + 8 = 0 y = 2^x 4y² - 33y + 8 = 0 delta d² = 33² - 4*4*8 = 961 d = 31 y1 = (33 + 31)/8 = 8 --> x1 = 3 y² = (33 - 31)/8 = 1/4 --> x2 = -2 pronto

ellencristinad Nada.. rsrsrs

luky007 4^(x+1) - 33.2^x + 8 = 0
2^(2x+2) - 33.2^x + 8 = 0
4*2^2x - 33.2^x + 8 = 0

y = 2^x

4y² - 33y + 8 = 0

delta
d² = - 4*4*8 = 961
d = 31
y¹ = (33 + 31)/8 = 8 --> x1 = 3
y² = (33 - 31)/8 = 1/4 --> x2 = -2

prontinho
espero ter ajudado

1 votes Thanks 4

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