Trouble manipulating a log expression This question sort of follows on from question Functions with logarithmic integrals. The book presents an example of integrating a function whose integral is logarithmic:
$$\int \frac{1}{4-3x} dx = -\frac{1}{3}\ln{|4 - 3x|} + K$$
$$= -\frac{1}{3}\ln{A|4 - 3x|}$$
$$= \frac{1}{3}\ln{\frac{A}{|4 - 3x|}}$$
I'm having trouble seeing how the final step is reached. My approach is to separate the logarithm of the product to the addition of separate logs then distribute the minus:
$$-\frac{1}{3}\ln{A|4 - 3x|} = -\frac{1}{3}(\ln{A} + \ln{|4 - 3x|}) = \frac{1}{3}(-\ln{A} - \ln{|4 - 3x|})$$
The I use the property that log minus another log is the log of the first divided by the second to get this:
$$\frac{1}{3}(-\ln{\frac{A}{|4-3x|}})$$
But I still have a minus that the example in the book doesn't have. Could someone help me with this please? Apologies for asking another question so soon.

From $-\dfrac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) $ we don't
get $\dfrac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }$, because 
$$\begin{equation*}
-\frac{1}{3}\ln \left( A\left\vert 4-3x\right\vert \right) \neq \frac{1}{3}
\ln \frac{A}{\left\vert 4-3x\right\vert }.
\end{equation*}$$
However if we write the constant of integration $C$ as $C=\frac{1}{3}\ln A$, we get the final result, as follows:
$$\begin{eqnarray*}
\int \frac{1}{4-3x}dx &=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +C \\
&=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +\frac{1}{3}\ln A \\
&=&\frac{1}{3}\left( -\ln \left\vert 4-3x\right\vert +\ln A\right)  \\
&=&\frac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }.
\end{eqnarray*}$$
