Question

Alguém me ajuda por favor
a) log10
b) log16 64
c) log1/9 √27
d) log1/4 4 + log4 1/4
e) log1/2 (log10 000)

Answer 1

a) log10 10 = x
10^x = 10^1
x = 1 ✓

b) log16 64 = x
16^x = 64
2^4x = 2^6
4x = 6
x = 6/4 = 3/2 ✓

c) log1/9 √27 = x
(1/9)^x = √27
(1/3)^2x = 27^1/2
1/3^2x = (3³)^1/2
3^(-2x) = 3^(3/2)
-2x = 3/2
-6x = 3
x = -6/3 = -1/2 ✓

d) log1/4 4 + log4 1/4
log1/4 1/4^(-1) + log4 4^(-1)
(-1) + (-1) = -2 ✓

e) log1/2 (log10 000)
Log1/2 ( Log 10^(4) )
log 1/2 ( 4 ) = x
(1/2)^x = 4
2^(-x) = 2^(2)
-x = 2
x = -2 ✓