Question

Resolva em IR as equações;a) Log₂(x-2) + 2Log₄ x=3LOg₈ (2x)
b) Log(3x) + Log(x-2)=Log(x+6) 

Answer 1

a)
C.E.:
x - 2 > 0  -->  x > 2
x > 0
2x > 0  -->  x > 0


$log_2(x-2)+2log_4x=3log_8(2x) \\ \\ log_2(x-2)+ \frac{2log_2x}{log_24} = \frac{3log_2(2x)}{log_28} \\ \\ log_2(x-2)+ \frac{2log_2x}{2}= \frac{3log_2(2x)}{3} \\ \\ log_2(x-2)+log_2x=log_2(2x) \\ \\ log_2[(x-2)*x]=log_2(2x) \\ \\ log_2[x^2-2x]=log_2(2x) \\ \\ x^2-2x=2x \\ \\ x^2-2x-2x=0 \\ \\ x^2-4x=0 \\ \\ x(x-4)=0 \\ \\ \\ x'=0 \\ \\ \\ x''-4=0 \\ \\x''=4$

S = {4}


b)
C.E.:
3x > 0  -->  x > 0
x - 2 > 0  -->  x > 2
x + 6 > 0  -->  x > -6


$log(3x) + log(x-2)=log(x+6) \\ \\ log[3x(x-2)]=log(x+6) \\ \\ log[3x^2-6x]=log(x+6) \\ \\ 3x^2-6x=x+6 \\ \\ 3x^2-6x-x-6=0 \\ \\ 3x^2-7x-6=0 \\ \\ Delta=121 \\ \\ x'=3 \\ \\ x''= \frac{-2}{3}$

S = {3}