Remember, any $\log$ to the base $b$, such as $\log_b(y)$, is in its heart an exponent. Let us start, then, from your expression $\log_b\left(\frac{1-3x}{x}\right)=3.$ Raise $b$ to the powers we see on each side. We get $b^{\log_b\left(\frac{1-3x}{x}\right)}=b^3.$ The left-hand side simplifies greatly. We get $\frac{1-3x}{x}=b^3.$ The rest is elementary algebra. The above equation is (for $x\ne 0$) equivalent to $1-3x=b^3 x,$ which is an easily solved linear equation.
Comment: There was no need to do the preliminary manipulation. We are told that $\log_b(1-3x)=3+\log_b x.$ Raise $b$ to the power on the left-hand side, the right-hand side. We obtain $b^{\log_b(1-3x)}=b^{3+\log_b x}.$ By the "laws of logarithms" this yields $1-3x=b^3 x.$