Answer
Solution set = $\displaystyle \{\frac{2}{9}\}.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llll}
x+4 \gt 0, & and & 5x+1 \gt 0 & \\
x \gt -4 & & x\gt-1/5 &
\end{array}\right]$
$ x \gt -1/5 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log(x+4)-\log 2=\log(5x+1)\qquad $... LHS: quotient rule
$\displaystyle \log\frac{x+4}{2}=\log(5x+1)\quad $ ... $\log $ is one-to-one
$\displaystyle \frac{x+4}{2}=5x+1\quad $ ... /$\times 2$
$ x+4=10x+2$
$ x-10x=2-4$
$-9x=-2$
$ x=\displaystyle \frac{-2}{-9}=\frac{2}{9}\qquad $ ... satisfies (*)
Thus, the solution set = $\displaystyle \{\frac{2}{9}\}.$
