I am unsure about raising e to the natural log of an absolute value (simple differential equation example) I am unsure what happens to the absolute value when solving this equation. 
The original question was asking: 
Find a particular solution to 
$\frac{dy}{dx} = 2y; y(1) = 3$
I have simplified it to the stage of $ln|y| = 2x+C$
What I am unsure about is what happens when I raise both sides to $e$.
 What happens to the absolute value?
Thanks.
 A: You get $\pm y=e^C\cdot e^{2x}$. The constant sign of $y(x)$ gets absorbed into the constant to form a new constant $C_1=\pm e^C$, to get the general solution $y(x)=C_1e^{2x}$.

Added note: If you still feel uncomfortable with the absolute value and sign, then take the homogenous solution as a hint and use the separation ansatz to get a strict derivation. 
For example, with $y'=2y+f(x)$, insert $y(x)=e^{2x}u(x)$, which is perfectly permissible since $x\mapsto e^{2x}$ is a positive smooth function. This results after a short computation in $u'(x)=e^{-2x}f(x)$ or 
$$u(x)=C+\int_{x_0}^x e^{-2s}f(s)\,ds$$
where now $C=u(x_0)$ is an arbitrary integration constant, positive, negative or zero, and then
$$y(x)=Ce^{2x}+\int_{x_0}^x e^{2(x-s)}f(s)\,ds=y(x_0)e^{2(x-x_0)}+\int_{x_0}^x e^{2(x-s)}f(s)\,ds$$
completely avoiding all discussion of $\ln|y|$.
A: $|y|$ is just a real number, which means that to begin with, if $\ln |y| = 2x + C$, it certainly follows that $|y| = e^{2x + C}$.  There is no problem so far.
The confusing part comes after taking $\exp$ to both sides, if you want to eliminate the absolute value signs from $y$.  To do so, you need a bit of a more sophisticated argument.  Essentially you say $y = \pm e^{2x + C}$ for each $x$, and then by continuity it will need to be the same $+$ or $-$ for all $x$.
As LutzL says, we often set $A = \pm e^C$ so that the solution becomes $y = A e^{2x}$ for some $A \in \mathbb{R}$.
A: HINT:
Substitute $| y| = a > 0$ and then we have:
$$\ln | y | = \ln a = 2x + C$$
$$e^{\ln a} = e^{2x + C}$$
$$a = |y| = e^{2x + C}$$
A: The condition $y(1)=3$ will decide the sign of $y$: only one sign is possible. You can take exponents to get $|y|=\exp(2x+C)$. Because $\exp$ is always positive, the condition $y(1)=3$ will determine the sign of $y$.
Another way of doing it is to write the constant as $\ln D$ rather than $C$ and say you'll let the condition $y(1)=3$ determine the sign of D. Then you have
$$\begin{align}
\ln y &= 2x + \ln D\\
y &= De^{2x}
\end{align}$$
Now plug in $y(1)=3$ to get your solution.
