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.

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what does $lnA[4-3*x]$ means?can you determine by words?is it product of A and 4-3*x? – dato datuashvili May 6 '12 at 11:32
It could be the case that A in the second step and the final step are not the same. But because it's an integration constant, it is kept as is. So the answer is 1/3*C/|4-3x| where C = 1/A. More like K = + 1/3*ln(A) – zubinmehta May 6 '12 at 11:33

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: $$\begin{eqnarray*} \int \frac{1}{4-3x}dx &=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +C \\ &=&-\frac{1}{3}\ln \left\vert 4-3x\right\vert +\frac{1}{3}\ln A \\ &=&\frac{1}{3}\left( -\ln \left\vert 4-3x\right\vert +\ln A\right) \\ &=&\frac{1}{3}\ln \frac{A}{\left\vert 4-3x\right\vert }. \end{eqnarray*}$$

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right because last one is equal to $1/3ln(A)-1/3ln(4-3*x)$ and by using rule of logarithms substraction,great answer @Américo Tavares – dato datuashvili May 6 '12 at 11:46
@dato: Thanks. Yes, by the quotient rule (or subtraction rule). – Américo Tavares May 6 '12 at 11:48
@AmericoTaveres may i ask one question?can i use this site problemasteoremas.wordpress.com in english language?because i dont know unfortunately spain – dato datuashvili May 6 '12 at 12:01
@dato My blog problemasteoremas.wordpress.com has only a few posts in English (search for the 'Math' tag). It is mainly in Portuguese. – Américo Tavares May 6 '12 at 12:58
ok thanks @ Américo Tavares – dato datuashvili May 6 '12 at 13:47

generally it would be clear that this one

$$\int \frac{1}{4-3x} dx = -\frac{1}{3}\ln{|4 - 3x|} + K$$ can be transformed into

$-ln[4-3*x]+3*K$ or
$ln[1/(4-3*x)]+3*K$

i hope this would help you,you should know that in case of $a*ln(f(x))$,$a$ comes in place of power Edited: because i gave answer into different interpretation,let denoted $3*K=ln(A)$,so now by using multiplicatipn rule,$ln(1/(4-3*x))+ln(A)=ln(A/(4-3*x))$

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