Question

responder as questoes 63, 64 e 65

Answer 1

Ola Islane 

63)

a) f(x) = √log2(x - 3) ⇒ Df(x) =  { R : x >= 4 }

log2(x - 3) ≥ 0 
log2(x - 3)≥ log2(1) 
x - 3 ≥ 1
x ≥ 4 

b) g(x) = 1/log1/2(x + 4) ⇒ Df(x) = { R : x > -3 or -4 < x < -3 }

log1/2(x + 4) > 0 
log1/2(x + 4) > log1/2)(1)
x + 4 > 1
x > -3 

x + 4 > 0 
x > -4 


c) h(x) = x/√log(2x) ⇒ Df(x) = { R : 2x > 1 }

√log(2x) > 0 
log(2x) > 0 
log(2x) > log(1) 
2x > 1

64) 

a) 

y = log3(x) 

y² - 3 ≥ 2y 

y² - 2y - 3 ≥ 0

delta
d² = 4 + 12 = 16
d = 4

y1 = (2 + 4)/2 = 3
y2 = (2 - 4)/2 = -1

y1 = log3(x)
log3(x1) = 3
x1 = 27 

y2 = log3(x2)
log3(x2) = -1
x2 = 1/3

x ≥ 27
0 < x ≤ 1/3

b) 

y = log1/2(x) 

y² - 3y - 4 > 0

delta 
d² = 9 + 16 = 25
d = 5

y1 = (3 + 5)/2 = 4
y2 = (3 - 5)/2 = -1 

y1 = log1/2(x1) = 4 ⇒ x1 = 1/16 
y2 = log1/2(x2) = -1 ⇒ x2 = 2

x > 2
0 < x < 1/16 

c)

log2(x)² < 4 

log2(x)² - 4 < 0

(log2(x) + 2)*(log2(x) - 2) < 0

log2(x) < -2 , x = 1/4

log2(x) = 2, x = 4

1/4 < x < 4 

65) 

-x² + log3(m)x - 1/4 = 0 

a) m = 9

-x² + log3(9)x - 1/4 = 0 

-x² + 2x - 1/4 = 0 

x² - 2x + 1/4 = 0 

delta
d² = 4 - 1 = 3
d = √3

x1 = (2 + √3)/2
x2 = (2 - √3)/2 

b) x² - log3(m)x + 1/4 = 0 

delta
d² = (log3(m))² - 1 > 0

(log3(m))²  > 1

m > 3 ,  0 < m < 1/3