Question

Se log2 (log 0,5 x) = 0 então x é igual a:

Answer 1

1. Se x = log27(base) 169 e y = log3(base) 13, entao:
a) x=2y/3
b) x= 3y
c) x= y/3
d) n.d.a

2. Se log 2 = 0,30 e log 3 = 0,48, o valor de log2(base) 3 eh:
a) 1,6
b) 0,8
c) 0,625
d) 0,5
e) 0,275

3. Se log8(base)x = m e x>0, entao log4(base)x eh igual a :
a) m/2
b) 3m/4
c) 3m/2
d) 2m
e) 3m

4. Se x= log8(base)25 e y= log2(base)5, entao:
a) x=y
b) 2x = y
c) 3x = 2y
d) x = 2y
e) 2y = 3y

5. O logaritmo de um numero na base 16 eh 2/3. Entao, o logaritmo desse numero na base 1/4 eh:
a) -4/3
b) -3/4
c) 3/8
d) 3
e) 6

6. Se x= log4(base)7 e y=log16(base)49, entao x-y eh:
a) log4(base)7
b) log16(base)7
c) 1
d) 2
e) 0

7. Sendo loga(base)2= x e loga(base)3= y, o valor de (log2(base)a + log3(base)a) . loga(base)4 . loga(base) raiz de 3 eh:
a) 2x + 2y
b) -2x - 2y
c) -x - y
d) x + y
e) x - y

8. Se log 2= 0,301, o valor de log100(base)1280 eh:
a) 1,0535
b) 1,107
c)1,3535
d) 1,5535
e) 2,107

x = log27(base) 169
x = log169 / log 27
x = log(13²) / log(3³)
x = 2 . log 13 / (3 . log 3)
x = (2/3) . log 13 / log 3
x = (2/3) . log3(base) 13

Substituindo log3(base) 13 por y:
x = (2/3) . y
x = 2y/3
--------------------------------------...
2.)
log 2 (base) 3 = log 3 / log 2 = 0,48 / 0,3 = 1,6
--------------------------------------...
3.)
log4(base)x =
log8(base)x / log8(base)4

Substituindo:
log8(base)x = m
log8(base)4 = 2/3

Então:
m / (2/3)
m . (3/2)
3m/2
--------------------------------------...
4.)
x= log8(base)25
x = log 25 / log 8
x = log (5²) / log (2³)
x = 2 . log 5 / (3 . log 2)
x = (2 / 3) . (log 5 / log 2)
x = (2 / 3) . log2(base)5

Substituindo log2(base)5 por y
x = (2 / 3) . y
x = 2y / 3
3 . x = 2y
3x = 2y
--------------------------------------...
5.)
número => n

log16(base)n = 2/3

log(1/4)(base)n = log16(base)n / log16(base)1/4
log(1/4)(base)n = 2/3 / (- 1/2)
log(1/4)(base)n = 2/3 . (- 2/1)
log(1/4)(base)n = - 4/3
--------------------------------------...
6.)
y = log16(base)49
y = log 49 / log 16
y = log (7²) / log (4²)
y = (2 . log 7) / (2 . log 4)
y = (2/2) . (log 7 / log 4)
y = (log 7 / log 4)
y = log4(base)7

Substituindo log4(base)7 por x
y = x
x = y
x - y = 0
--------------------------------------...
7.)
Sendo:
x = log a (base) 2
y = log a (base) 3

Então:
1/x = log 2 (base) a
1/y = log 3 (base) a

(log 2 (base) a + log 3 (base) a) . log a (base) 4 . log a (base) raiz de 3

(log 2 (base) a + log 3 (base) a) . log a (base) 2² . log a (base) 3^(1/2)

(log 2 (base) a + log 3 (base) a ) . 2 . log a (base) 2 . 1/2 . log a (base) 3

(log 2 (base) a + log 3 (base) a) . log a (base) 2 . log a (base) 3

Substituindo:
(1/x + 1/y). x . y
(1/x + 1/y). xy
(xy)/x + (xy)/y
y + x
x + y
--------------------------------------...
8.)
A questão nº 8 pode ser resolvida de duas maneiras:

1ª maneira
log 2 = 0,301

log100(base)1280 = log2(base)1280 / log2(base)100
log100(base)1280 = log2(base)(128.10) / log2(base)10²
log100(base)1280 =
[log2(base)128 + log2(base) 10] / 2.[log2(base)10]
[7 + 1/0,301] / 2.[1/0,301]
[7 + 1/0,301] / [2/0,301]
[7 + 1/0,301] . [0,301/2]
[7 + 1/0,301] . [0,1505]
(7 . 0,1505) + (1/0,301 . 0,1505)
1,0535 + (0,1505/0,301)
1,0535 + 0,5
1,5535



2ª maneira (a mais fácil)
log 2 = 0,301

log100(base)1280 = log 1280 / log 100
log100(base)1280 = log (128 . 10) / log 10²
log100(base)1280 = log (2^7 . 10) / log 10²
log100(base)1280 = [log (2^7) + log 10] / log 10²
log100(base)1280 = (7 . log 2 + log 10) / (2 . log 10)

Sabendo que:
log 10 = 1
log100(base)1280 = (7 . log 2 + 1) / (2 . 1)
log100(base)1280 = (7 . log 2 + 1) / 2
log100(base)1280 = (7 . 0,301 + 1) / 2
log100(base)1280 = (2,107 + 1) / 2
log100(base)1280 = 3,107 / 2
log100(base)1280 = 1,5535
--------------------------------------...
Lista de alternativas corretas:
1.) a
2.) a
3.) c
4.) c
5.) a
6.) e
7.) d
8.) d