Determine o valor de x: a) logx (x + 6) = 2 b) log2(3x + 2) = 5 c) log2 (3x − 3) = log2 9 d) log(x + 4) + log(x − 4) = 2 ∙ log 3
a) log x (x + 6) = 2
Means that
x^2 = x + 6
x^2 - x - 6 = 0
(x - 3) ( x+ 2) = 0
x - 3 = 0 x + 2 = 0
x = 3 x = - 2 ( reject )
b) log 2 (3x +2) = 5
Means that
2^5 = 3x + 2
32 = 3x + 2
32 - 2 = 3x
30 = 3x
10 = x
c) log 2 ( 3x - 3) = log 2 9
Implies that
3x - 3 = 9
3x = 9 + 3
3x = 12
x = 4
d) log ( x + 4) + log )x - 4) = 2 log 3
We can write
log [ ( x + 4) (x - 4) ] = log 3^2
So
(x + 4) ( x - 4) = 3^2
x^2 - 16 = 9
x^2 = 16 + 9
x^2 = 25 take the positive root
x = 5