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.