Answer
The solution is $\left\{ 2 \right\}$.
Work Step by Step
Solve the provided equation as follows:
$\begin{align}
& \log \left( x+3 \right)+\log x=1 \\
& log\left[ \left( x+3 \right)\left( 3 \right) \right]=1 \\
& \left( x+3 \right)x={{10}^{1}} \\
& {{x}^{2}}+3x=10
\end{align}$
Further solve this equation
$\begin{align}
& {{x}^{2}}+3x-10=0 \\
& {{x}^{2}}+5x-2x-10=0 \\
& x\left( x+5 \right)-2\left( x+5 \right)=0 \\
& \left( x+5 \right)\left( x-2 \right)=0
\end{align}$
Finally, we get, $ x=-5,2$
Put the value of $ x=-5$ in the original equation,
$\log \left( -5+3 \right)+\log \left( -5 \right)=1$
Since, log of a negative number is not defined, $ x=-5$ is not solution
Thus the solution set is $\left\{ 2 \right\}$.
