If an equation implies a single solution can this solution possibly be extraneous? I am going through a book where they keep verifying solutions even when there is only one solution to an equation. To me that seems unnecessary, but I don't have a clear explanation. I would like to confirm whether, in that case, the verification is indeed unnecessary–in which case I would love an explanation.
Example
$$
\log_3(2x-1)+\log_{1/3}(x+2)=0\\
\log_3(2x-1)=\log_{3}(x+2)\\
2x-1=x+2\\
x=3 \text{ (satisfies)}
$$
 A: One of the things that kind of bothers me about most of the precalculus texts out there is that they encourage students to plow ahead with computations, then check those computations by substitution at the end.  This leads to a procedural view of mathematics, and tends to hide understanding.  When I teach these kinds of problems, I typically take the following approach.

Presumably, we are regarding $\log_b$ as a real function, i.e. a function that takes positive real inputs and gives real outputs.  Thus $\log_b: (0,\infty)\to\mathbb{R}$, which says that the domain of $\log_b$ is the positive real numbers.  Hence if
$$ \log_3(2x-1) + \log_{\frac{1}{3}}(x+2) = 0 $$
is to have any solutions $x$, then those solutions must satisfy
$$ 2x - 1 > 0 \implies x > \frac{1}{2} \tag{1} $$
and
$$ x + 2 > 0 \implies x > -2. \tag{2} $$
Thus we know a priori that the only possible solutions must be greater than $\frac{1}{2}$.
Now, provided that both (1) and (2) hold, we can solve the equation as originally proposed.  That is, we have
\begin{align}
\log_3(2x-1) + \log_{\frac{1}{3}}(x+2) = 0
&\iff \log_3(2x-1) = \log_3(x+2) \\
&\iff 2x-1 = x+2 \\
&\iff x=3.
\end{align}
Notice that $x = 3 > \frac{1}{2}$, so this value of $x$ gives a real solution to the original equation.  We still have to check the legitimacy of this solution, but it comes down to checking that the solution is in the appropriate domain.

This may not seem like it saves us much work, but consider another example.  Suppose that we want to solve
$$ \log_3(-2x-6) + \log_{\frac{1}{3}}(x+2) = 0? $$
We need
$$ -2x-6 > 0 \implies x < -3
\qquad\text{and}\qquad
x+2 > 0 \implies x > -2. $$
This is clearly nonsense, since there is no number $x$ that is simultaneously less than $-3$ and greater than $-2$.  Without doing any additional steps, we know that there are no solutions.
However, if we attempted to solve as originally proposed, we would have
\begin{align}
\log_3(-2x-6) + \log_{\frac{1}{3}}(x+2) = 0
&\implies \log_3(-2x-6) = \log_3(x+2) \\
&\implies -2x-6 = x+2\\
&\implies -3x = 8 \\
&\implies x = -\frac{8}{3}.
\end{align}
We could then substitute this into the original equation and find out that it doesn't give a real solution, but this seems like a lot of extra work.

TL;DR version:  you always have to check for extraneous solutions if you plow ahead with computation before doing anything else.  However, if you first check that the question makes sense (by checking the domain), you can generally save yourself a lot of work and recognize extraneous solutions when they pop up.
A: You need to check the domain.
For example, try to solve the following equation.
$$\log_3(2x-1)+\log_{\frac{1}{3}}(x-2)=0.$$
