1. \(\displaystyle{8}^{{{x}}}-{15}={0}\)

\(\displaystyle{8}^{{{x}}}={15}\)

Take ln both side

\(\displaystyle{{\ln{{8}}}^{{{x}}}=}{\ln{{15}}}\)

Using, \(\displaystyle{{\ln{{a}}}^{{{b}}}=}{b}{\ln{{a}}}\)

\(\displaystyle{x}{\ln{{8}}}={\ln{{15}}}\)

\(\displaystyle{x}={\frac{{{\ln{{15}}}}}{{{\ln{{8}}}}}}\)

\(\displaystyle{x}={1.3023}\)

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

\(\displaystyle{\log{{\left({x}+{6}\right)}}}-{\log{{\left({3}{x}-{2}\right)}}}={2}\)

Using, \(\displaystyle{\log{{a}}}-{\log{{b}}}={\log{{\left({\frac{{{a}}}{{{b}}}}\right)}}}\)

\(\displaystyle{\log{{\left({\frac{{{x}+{6}}}{{{3}{x}-{2}}}}\right)}}}={2}\)

Using, \(\displaystyle{\log{{a}}}={b}\)

then \(\displaystyle{a}={10}^{{{b}}}\)

\(\displaystyle{\frac{{{x}+{6}}}{{{3}{x}-{2}}}}={10}^{{{2}}}\)

\(\displaystyle{x}+{6}={\left({3}{x}-{2}\right)}\times{100}\)

\(\displaystyle{x}+{6}={300}{x}-{200}\)

\(\displaystyle{300}{x}-{x}={200}+{6}\)

\(\displaystyle{299}{x}={206}\)

\(\displaystyle{x}={\frac{{{206}}}{{{299}}}}\)

\(\displaystyle{x}={0.6889}\)

3. \(\displaystyle{4400}={2000}{\left({1}+{\frac{{{0.045}}}{{{12}}}}\right)}\)

\(\displaystyle{\frac{{{4400}}}{{{2000}}}}={\left({\frac{{{12}+{0.045}}}{{{12}}}}\right)}^{{{12}{x}}}\)

\(\displaystyle{2.2}={\left({1.0038}\right)}^{{{12}{x}}}\)

Take ln both side

\(\displaystyle{\ln{{2.2}}}={{\ln{{\left({1.0038}\right)}}}^{{{12}{x}}}}\)

\(\displaystyle{\ln{{\left({2.2}\right)}}}={12}{x}{\ln{{\left({1.0038}\right)}}}\)

\(\displaystyle{x}={\frac{{{\ln{{2.2}}}}}{{{12}{\ln{{1.0038}}}}}}\)

\(\displaystyle{x}={17.3236}\).

\(\displaystyle{8}^{{{x}}}={15}\)

Take ln both side

\(\displaystyle{{\ln{{8}}}^{{{x}}}=}{\ln{{15}}}\)

Using, \(\displaystyle{{\ln{{a}}}^{{{b}}}=}{b}{\ln{{a}}}\)

\(\displaystyle{x}{\ln{{8}}}={\ln{{15}}}\)

\(\displaystyle{x}={\frac{{{\ln{{15}}}}}{{{\ln{{8}}}}}}\)

\(\displaystyle{x}={1.3023}\)

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

\(\displaystyle{\log{{\left({x}+{6}\right)}}}-{\log{{\left({3}{x}-{2}\right)}}}={2}\)

Using, \(\displaystyle{\log{{a}}}-{\log{{b}}}={\log{{\left({\frac{{{a}}}{{{b}}}}\right)}}}\)

\(\displaystyle{\log{{\left({\frac{{{x}+{6}}}{{{3}{x}-{2}}}}\right)}}}={2}\)

Using, \(\displaystyle{\log{{a}}}={b}\)

then \(\displaystyle{a}={10}^{{{b}}}\)

\(\displaystyle{\frac{{{x}+{6}}}{{{3}{x}-{2}}}}={10}^{{{2}}}\)

\(\displaystyle{x}+{6}={\left({3}{x}-{2}\right)}\times{100}\)

\(\displaystyle{x}+{6}={300}{x}-{200}\)

\(\displaystyle{300}{x}-{x}={200}+{6}\)

\(\displaystyle{299}{x}={206}\)

\(\displaystyle{x}={\frac{{{206}}}{{{299}}}}\)

\(\displaystyle{x}={0.6889}\)

3. \(\displaystyle{4400}={2000}{\left({1}+{\frac{{{0.045}}}{{{12}}}}\right)}\)

\(\displaystyle{\frac{{{4400}}}{{{2000}}}}={\left({\frac{{{12}+{0.045}}}{{{12}}}}\right)}^{{{12}{x}}}\)

\(\displaystyle{2.2}={\left({1.0038}\right)}^{{{12}{x}}}\)

Take ln both side

\(\displaystyle{\ln{{2.2}}}={{\ln{{\left({1.0038}\right)}}}^{{{12}{x}}}}\)

\(\displaystyle{\ln{{\left({2.2}\right)}}}={12}{x}{\ln{{\left({1.0038}\right)}}}\)

\(\displaystyle{x}={\frac{{{\ln{{2.2}}}}}{{{12}{\ln{{1.0038}}}}}}\)

\(\displaystyle{x}={17.3236}\).