Answer
Solution set = $\displaystyle \{\frac{1}{5}\}.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llll}
x+7 \gt 0, & and & 7x+1 \gt 0 & \\
x \gt -7 & & x\gt-1/7 &
\end{array}\right]$
$ x \gt -1/$7$ \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log(x+7)-\log 3=\log(7x+1) \qquad $... LHS: quotient rule
$\displaystyle \log\frac{x+7}{3}=\log(7x+1) \quad $ ... $\log $ is one-to-one
$\displaystyle \frac{x+7}{3}=7x+1 \quad $ ... /$\times $3
$ x+7=21x+3$
$7-3=21x-x $
$4=20x $
$ x=\displaystyle \frac{4}{20}=\frac{1}{5}\qquad $ ... satisfies (*)
Thus, the solution set = $\displaystyle \{\frac{1}{5}\}.$
