\(\displaystyle{8}={2}^{{3}},\ {s}{o}\ {\log{{8}}}={\left(\frac{{1}}{{3}}\right)}{\log{{2}}}\)

\(\displaystyle{\log{{2}}}{\left({x}+{2}\right)}-{\log{{2}}}{\left({x}+{3}\right)}={2}\)

\(\displaystyle{\log{{2}}}{\left[\frac{{{x}+{2}}}{{{x}+{3}}}\right]}={2}\)

thus, \(\displaystyle\frac{{{x}+{2}}}{{{x}+{3}}}={2}^{{2}}={4}\)

x+3=4*(x+3)

x+3=4x+12

3x=3-12=-9

x=-3