Solve each of the following equations for x. Determine the exact value for x and

Solve each of the following equations for x. Determine the exact value for x and a decimal approximation correct to four decimals places.
1. $$\displaystyle{8}{x}-{15}={0}$$(Solving this equation will locate the x-intercept on the gtaph of $$\displaystyle{y}={8}^{{{x}}}-{15}$$)
2. $$\displaystyle{\log{{\left({x}+{6}\right)}}}={\log{{\left({3}{x}-{2}\right)}}}+{2}$$ (Solving this equation will determine the x-coordinate of the point of intersection of the functions $$\displaystyle{y}={\log{{\left({x}+{6}\right)}}}$$ and $$\displaystyle{y}={\log{{\left({3}{x}-{2}\right)}}}+{2}.{)}$$
3. $$\displaystyle{4400}={2000}{\left({1}+{\frac{{{0.045}}}{{{12}}}}\right)}^{{{12}{x}}}$$ (Solving this equation will determine the amount of time it takes an investment $2000 to grow to$4400 if interest is $$\displaystyle{4.5}\%$$ compounded monthly.)

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Pohanginah
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
$$\displaystyle{\ln{{2.2}}}={{\ln{{\left({1.0038}\right)}}}^{{{12}{x}}}}$$
$$\displaystyle{\ln{{\left({2.2}\right)}}}={12}{x}{\ln{{\left({1.0038}\right)}}}$$
$$\displaystyle{x}={\frac{{{\ln{{2.2}}}}}{{{12}{\ln{{1.0038}}}}}}$$
$$\displaystyle{x}={17.3236}$$.