Answer
$x=\dfrac{e^6}{2}$
Work Step by Step
Using the properties of logarithms, the given equation, $
\ln x + \ln2=6
,$ is equivalent to
\begin{align*}\require{cancel}
\ln [x(2)]&=6
&\left(\text{use } \ln (ab)=\ln a+\ln b \right)
\\
\ln (2x)&=6
\\
\log_e (2x)&=6
&\left(\text{use } \ln x=\log_ex \right)
\\
(2x)&=e^6
&\left(\text{use } y=b^x\Rightarrow\log_by=x \right)
\\\\
\dfrac{(2x)}{2}&=\dfrac{e^6}{2}
\\\\
x&=\dfrac{e^6}{2}
.\end{align*}
Hence, the solution is $
x=\dfrac{e^6}{2}
.$
CHECKING: Substituting $x=\dfrac{e^6}{2}$ in the given equation, $\ln x + \ln2=6,$ results to
\begin{align*}
\ln \left(\dfrac{e^6}{2}\right) + \ln2&=6
\\\\
\ln e^6-\ln2+\ln2&=6
&\left(\text{use } \ln \dfrac{a}{b}=\ln a-\ln b \right)
\\\\
\ln e^6-\cancel{\ln2}+\cancel{\ln2}&=6
\\\\
\ln e^6&=6
\\\\
6(\ln e)&=6
&\left(\text{use } \ln a^x=x\ln a \right)
\\\\
6(1)&=6
&\left(\text{use } \ln e=1 \right)
\\
6&=6
&\text{ (TRUE)}
.\end{align*}
Since the substitution above ended with a TRUE statement, then $
x=\dfrac{e^6}{2}
$ is the solution.