TITLE: Why is this constant of integration taken as $\log A$ instead of just $C$?
QUESTION [8 upvotes]: Suppose we solve 
$$\frac{dy}{dx} = \frac{1 + y}{2 + x} .$$
Which can be written as the following and integrating both sides w.r.t. $y$ and $x$: $$\int\frac{1}{1 + y}dy = \int\frac{1}{2 +x}dx ,$$
we get $$\ln(1+y) = \ln(2+x) + C$$ 
One of the book says: 

It's convenient to write the constant $C$ as the logarithm of some other constant $A$:
  $$ \ln(1+y) = \ln(2+x) + \ln(A) \implies \ln A(2 + x)$$
  $$ \therefore (1 + y) = A(2 + x)$$ 

Question: Why is it "convenient to write the constant $C$ as the logarithm of some other constant $A$"? What liberty do we have to write $\ln(A)$ instead of just $C$? I think I am unaware of what a logarithm of a constant is. I mean the significance of it.

REPLY [7 votes]: It is convenient because it allows you to write both sides of an equation as a logarithm, and then use the fact that $\ln(a)=\ln(b)$ implies $a=b$.
You have liberty to do so because $\ln$ is a surjective function, meaning that every real number has the form $\ln(A)$ for some $A>0$.
However, it is also unnecessary to do so, so do not worry if this style of argument seems unnatural at present.  From $\ln(1+y)=\ln(2+x)+C$, you can exponentiate to get $1+y=e^C(2+x)$, and then recognize that $e^C$ is also an arbitrary (but positive) constant, and you might as well call it $A$.
A side note: $\ln(1+y)=\ln(2+x)+C$ is not quite a valid deduction, unless you are assuming that $y>-1$ and $x>-2$.

REPLY [3 votes]: As I said in my answer to your last question, what we need when solving a differential equation (a simple example of which is taking a definite integral) is a constant of integration that can take on any real value. For this reason, we cannot write $A^2$ as our constant of integration, because we can never get a negative value for $A^2$ by plugging in a real value for $A$. However, we can write $\ln(A)$ for our constant because for any real number $r$ we get our constant of integration equal to $r$ when we plug in $e^r$ for $A$. 
As for why using $\ln(A)$ as the constant of integration is more convenient than using $C$, its just so that you can write $(1+y)=A(2+x)$ rather than $(1+y)=e^C(2+x)$, and whoever wrote your textbook decided the first was a nicer expression.