Answer
Solution set = $\{2\}.$
Work Step by Step
For the equation to be defined, logarithms must have positive arguments
$\left[\begin{array}{llll}
x \gt 0, & and & x+3 \gt 0 & \\
& & x\gt-3 &
\end{array}\right]$
$ x \gt -3 \qquad (*)$
This is the condition eventual solutions must satisfy.
$\log x+\log(x+3)=\log 10\qquad $... LHS: product rule
$\log(x^{2}+3x)=\log 10 \quad $ ... $\log $ is one-to-one
$ x^{2}+3x=10$
$ x^{2}+3x-10=0$
Factor: +5 and -2 are factors of -10 whose sum is +3.
$(x+5)(x-2)=0$
$ x=2 \qquad $ ... satisfies (*)
$ x=-5 \quad $ ... does not satisfy (*)
Thus, the solution set = $\{2\}.$