Question

b) log 4x + log4 (x+3)=1
c) log5 (2x/x+1)= log 5 -1
d) log 4 2x + log 4 (9-x)= 2
e) log (x 2 -1)(x elevado a 2) = log (2x-1)

Answer 1

Ola Robson

b) 

log4(x) + log4(x + 3) = 1 = log4(4)

x*(x + 3) = 4

x² + 3x - 4 = 0

delta
d² = 9 + 16 = 25
d = 5

x = (-3 + 5)/2 = 2/2 = 1

c)

log5(2x/(x + 1)) = log5(-1) 

2x/(x + 1) = -1

2x = -x - 1
3x = -1
x = -1/3 

não existe solução real

d)

log4(2x) + log4(9 - x) = 2

log4(2x) + log4(9 - x) = log4(16) 

2x*(9 - x) = 16 

2x² - 18x + 16 = 0

x² - 9x + 8 = 0

delta
d² = 81 - 32 = 49
d = 7

x1 = (9 + 7)/2 = 8
x2 = (9 - 7)/2 = 1

e)

log(x² - 1) = log(2x - 1) 

log(x² - 1) - log(2x - 1) = 0 

log( (x² - 1)/(2x - 1) ) = log(1)

x² - 1 = 2x - 1

x² - 2x = 0

x*(x - 2) = 0

x = 2