Answer
$\left\{-2,-1\right\}$
Work Step by Step
Since $\ln x=\ln y$ implies $x=y$, then the given equation, $
\ln(x^2+3x+4)=\ln2
$, implies
\begin{align*}
x^2+3x+4&=2
.\end{align*}
Using the properties of equality, the equation above is equivalent to
\begin{align*}\require{cancel}
x^2+3x+4-2&=0
\\
x^2+3x+2&=0
.\end{align*}
Using the factoring of trinomials, the equation above is equivalent to
\begin{align*}\require{cancel}
(x+2)(x+1)&=0
.\end{align*}
Equating each factor to zero (Zero Product Property) and solving for the variable, then
\begin{array}{l|r}
x+2=0 & x+1=0
\\
x=-2 & x=-1
.\end{array}
Hence, the solution set of the equation $
\ln(x^2+3x+4)=\ln2
$ is $
\left\{-2,-1\right\}
$.
