Translation Invariant of Laplace equation The Laplace equation$$\Delta u = 0$$ is translation invariant. However, the radial 2-D solution $$u(x)=log(r)*a$$ is obviously not translation invariant. So is it formally correct to write $\Delta u = \delta_0(x)$ as in this form the equation is no more translation invariant. Why there are still people writing $\Delta u=0$?
 A: I will write something in this answer to avoid extended discussion in comments. It is a bit lengthy but I hope that it is an easy read. 

Let us reason formally and consider an equation 
$$\tag{1}
F(u)=0, $$ 
where $u$ belongs to some function space $U$ and $F\colon U\to U$ (not necessarily a linear operator). We also have a group $G$ acting on $U$ and we write $g\cdot u$ to denote such action; if $G=(\mathbb R^n, +)$ is the translation group, then $(g\cdot u)(x)=u(x-g)$. We say that (1) is invariant under the action of $G$ (or $G$-invariant) if $g\cdot u$ is a solution to (1) whenever $u$ is. A sufficient (not necessary) condition for this is the following: 
$$\tag{2}
F(g\cdot u)=g\cdot F(u),\quad \forall u\in U,\ g\in G.$$
This is exactly the case of $F=-\Delta$ and $G=(\mathbb R^n, +)$. 

You asked if (2) implies that solutions $u$ to (1) are $G$ invariant, that is, $g\cdot u = u$ for all $g\in G$. The answer is affirmative if one knows a priori that the solution to (1) is unique. In that case, if $u$ is the solution to (1), then $g\cdot u$ must also be a solution and so $g\cdot u=u$. For example, let 
$$\Omega=\{ (x, y)\in \mathbb R^2\ :\ 1<x^2+y^2<2\}$$
and 
$$U=\left\{f\colon \Omega \to \mathbb R\ :\ \lim_{|\mathbf x|\to 2^-}f(\mathbf x) = a, \lim_{|\mathbf x|\to 1^+} f(\mathbf x) = b\right\}$$ 
for constants $a, b$. Now, if $G=O(2)$ and $F=-\Delta$, then both $F$ and $U$ are $G$-invariant and since $\Omega$ is a bounded and smooth domain, the solution to (1) is unique (you can see this via the energy integral). Therefore we can conclude that this unique solution must be $G$-invariant, that is, the unique solution is radially symmetric. 

However, as you noticed in comments to the main question, people still look for $G$-invariant solutions to (1) even if no uniqueness result hold. Why? One reason is simply that they might have no idea of how a solution might look like, and so they begin with the most symmetric ansatz to see if they are lucky. Another reason is that, in the linear case, they might be on the hunt for a fundamental solution, that is, a solution $E$ to $F(E)=\delta$. Once they have just one fundamental solution (usually there are many), they can solve $F(u)=g$ for virtually all $g$. It makes sense, then, to make the most symmetric possible ansatz, to make the fundamental solution as simple as possible. 
The most systematic exploitation of these tricks is the technique of separation of variables. For this I recommend having just a look at the book Symmetry and separation of variables, by W.Miller. 

Finally, there is still another viewpoint that applies to the linear case. Suppose that you know an orthonormal basis $e_{\xi}$ of $U$, indexed by $\xi$, such that the action of $G$ on $e_{\xi}$ is the simplest possible: 
$$g\cdot e_\xi = k_\xi(g)e_\xi, $$ 
where $k_\xi(g)\in \mathbb C\setminus\{0\}$ (this needs that $G$ be Abelian, but the method can be generalized to non-Abelian groups as well). If $F$ is a linear operator that satisfies (2), an amazingly simple and deep result known as Schur's lemma of representation theory implies that $F$ is diagonalized by $e_\xi$, that is 
$$F(e_\xi)=m(\xi) e_\xi.$$ 
Therefore, if one decomposes the unknown function as
$$\tag{3}u=\sum_\xi \hat{u}(\xi)e_\xi, $$ 
one has, by linearity, that $F(u)=\sum_\xi m(\xi)\hat{u}(\xi)e_\xi, $ which allows to transform equations such as (1) and its nonhomoegeneous counterparts into algebraic equations. This is the viewpoint of Fourier analysis and representation theory. 
For the equation in the OP, where $F=-\Delta$ and $G=(\mathbb R^n, +)$, one has that $$e_\xi(x)=\exp(i x\cdot \xi),\qquad \xi\in \mathbb R^n$$ and $$m_\xi=|\xi|^2.$$  Since $\xi$ runs over a continuum, the decomposition in (3) is given by an integral, rather than a sum:
$$u(x)=\int_{\mathbb R^n} \hat{u}(\xi) \exp(i x\cdot \xi)\, \frac{d\xi}{(2\pi)^n}.$$ This is the Fourier inversion formula. To see how this is applied to the analysis of the Laplace (and Poisson) differential equation, I like the book "Real Analysis" by Gerry Folland, 1st or 2nd edition, chapter "Elements of Fourier analysis" and section "Application to partial differential equations" (in 2nd edition this is §8.7).
A: 
Why there are still people writing $\Delta u=0$ 

Maybe because they have context around that formula, such as "$\Delta u=0$ except at $0$", or "$\Delta u = 0$ in the domain of $u$".  It would be incorrect to say that $\log r$ satisfies the equation $\Delta u=0$ in the entire plane. 
Whether it's correct to write $\Delta u = \delta_0(x)$  depends on the context, too. Maybe the equation is not considered in the sense of distributions at all, it's just considered in the classical sense in the domain $\mathbb{R}^2\setminus \{0\}$.  In any event, there would be a multiple of $2\pi$ with the delta-function. 
The discussion of translation invariance seems confused. A function being translation invariant means $f(x+h)=f(x)$ where $h$ is the vector of translation. A particular solution of a translation-invariant equation does not have to be a translation-invariant function. For example, the equation $\frac{dy}{dx}  =  5$ is translation invariant, but its solution $f(x) = 5x+2$ is not translation invariant, as $f(x+h)$ is not equal to $f(x)$.  What is true is that the translation of a solution is also a solution, but possibly a different one.
