3. For each value of $a$, find all values of $x$ that satisfy the equation
$$
\begin{array}{l}
\log _{5}\left[\frac{(x+1)^{2}}{x}-a\right] \\
=\log _{5} \frac{(x+1)^{2}}{x}-\log _{5} a
\end{array}
$$
3. Notice that
$$
\begin{array}{l}
\log _{5}\left[\frac{(x+1)^{2}}{x}-a\right] \\
=\log _{5} \frac{(x+1)^{2}}{x}-\log _{5} a \\
\Leftrightarrow\left\{\begin{array}{l}
\log _{5}\left[\frac{(x+1)^{2}}{x}-a\right]=\log _{5} \frac{(x+1)^{2}}{a x}, \\
a>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
\frac{(x+1)^{2}}{x}-a=\frac{(x+1)^{2}}{a x} \\
\frac{(x+1)^{2}}{x}>0, a>0
\end{array}\right. \\
\Leftrightarrow\left\{\begin{array}{l}
x>0, a>0, \\
(a-1)(x+1)^{2}=a^{2} x .
\end{array}\right.
\end{array}
$$

If $a=b$, then $x=0$, which contradicts $x>0$.
If $a \neq 1$, then
$$
\frac{(x+1)^{2}}{x}=\frac{a^{2}}{a-1} \text {. }
$$

Solving this, we get $x=a-1$ or $\frac{1}{a-1}$.
Also, by $x>0$, we get $a>1$.
Therefore, when $a>1$, $x=a-1$ or $\frac{1}{a-1}$;
when $a \leqslant 1$, the equation has no solution.