We are given
\(\displaystyle{\mid}{\ln{{\left({x}+{3}{\mid}={1}\right.}}}\)

Since the left side is an absolute value expression and the right side is positive, there are two possible equations:

\(\displaystyle{\ln{{\left({x}+{3}\right)}}}=-{1}\ {\ln{{\left({x}+{3}\right)}}}={1}\)

\(\displaystyle{x}+{3}={e}^{{-{{1}}}}\ {x}+{3}={c}^{{1}}\)

\(\displaystyle{x}+{3}=\frac{{1}}{{c}}\ {x}+{3}={e}\)

\(\displaystyle{x}=\frac{{1}}{{c}}-{3}\ {x}={e}-{3}\)

\(\displaystyle{x}=\frac{{{1}-{3}{c}}}{{e}}\)

So, the solutions are:

\(\displaystyle{x}=\frac{{{1}-{3}{e}}}{{e}},{x}={e}-{3}\)

Since the left side is an absolute value expression and the right side is positive, there are two possible equations:

\(\displaystyle{\ln{{\left({x}+{3}\right)}}}=-{1}\ {\ln{{\left({x}+{3}\right)}}}={1}\)

\(\displaystyle{x}+{3}={e}^{{-{{1}}}}\ {x}+{3}={c}^{{1}}\)

\(\displaystyle{x}+{3}=\frac{{1}}{{c}}\ {x}+{3}={e}\)

\(\displaystyle{x}=\frac{{1}}{{c}}-{3}\ {x}={e}-{3}\)

\(\displaystyle{x}=\frac{{{1}-{3}{c}}}{{e}}\)

So, the solutions are:

\(\displaystyle{x}=\frac{{{1}-{3}{e}}}{{e}},{x}={e}-{3}\)