Question

2. Seja f(x) = Log3 (3x+4) – Log3(2x-1). Os valores de X, para os quais f está
definida e satisfaz f(x) > 1, são:

a) x< 7/3
b) 1/2< x
c) 1/2< x < 7/3
d) -4/3 < x
e) -4/3 < x < 1/2

Answer 1

Resposta:

c)

Explicação passo-a-passo:

$Lembrando\ da\ propriedade\ dos\ logaritimos:\\log_ab - log_ac = log_a\frac{b}{c} \\\\temos\ entao\ que:\\f(x) = log_3(3x+4)-log_3(2x-1) = log_3\frac{3x+4}{2x-1}$

$a\ funcao\ log_a(x)\ vale\ para\ todo\ x>0\\entao\ f(x)\ vale\ para\ \frac{3x+4}{2x-1} >0\\\\estudo\ do\ sinal:\\3x+4>0 \rightarrow x>-\frac{4}{3} ,\ 3x+4 = 0 \rightarrow x=-\frac{4}{3}, \ 3x+4 < 0 \rightarrow x<-\frac{4}{3} \\2x-1>0 \rightarrow x>\frac{1}{2} ,\ 2x-1 = 0 \rightarrow x=\frac{1}{2}, \ 2x-1 < 0 \rightarrow x<\frac{1}{2} \\\\\frac{3x+4}{2x-1} >0 \rightarrow x> \frac{1}{2},\ e\ x < -\frac{4}{3}$

$f(x) >1 \rightarrow log_3\frac{3x+4}{2x-1} >1 \rightarrow \frac{3x+4}{2x-1} >3^1 \rightarrow \frac{3x+4}{2x-1} - 3 >0 \rightarrow \frac{-3x+7}{2x-1} >0 \\\\estudo\ do\ sinal:\\-3x+7>0 \rightarrow x< \frac{7}{3},\ -3x+7=0 \rightarrow x = \frac{7}{3},\ -3x+7<0 \rightarrow x> \frac{7}{3}\\2x-1>0 \rightarrow x> \frac{1}{2},\ 2x-1=0 \rightarrow x = \frac{1}{2},\ 2x-1<0 \rightarrow x< \frac{1}{2}\\\\\frac{-3x+7}{2x-1} > 0 \rightarrow \frac{1}{2} <x<\frac{7}{3}$