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
$$1-3x=b^3 x.$$