[tex]\log_{10}x=\log x[/tex]
[tex]The\ domain:\\3x-1 > 0\to x > \dfrac{1}{3}\to x\in\left(\dfrac{1}{3},\ \infty\right)\\\\\log(3x-1)-\log2=3\ \ \ \ \boxed{\text{use}\ \log_a\dfrac{b}{c}=\log_ab-\log_ac}\\\\\log\dfrac{3x-1}{2}=3\ \ \ \ \boxed{\text{use}\ \log_ab=c\iff a^c=b}\\\\\dfrac{3x-1}{2}=10^3\\\\\dfrac{3x-1}{2}=1000\ \ \ \ \ |\cdot2\\\\3x-1=2000\ \ \ \ |+1\\\\3x=2001\ \ \ \ |:3\\\\x=667 > \dfrac{1}{3}\\\\Answer:\ \boxed{x=667}[/tex]
Log10(3x-1)-log10(2) = 3
log10 [3x-1)/2 = 3
(3x - 1) / 2 = 10^3
3x - 1 = 2000
x = 2001 / 3
= 667 Answer
