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

Efan Halliday 2021-09-16 Answered

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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Expert Answer

Pohanginah
Answered 2021-09-17 Author has 21438 answers
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
\(\displaystyle{\ln{{2.2}}}={{\ln{{\left({1.0038}\right)}}}^{{{12}{x}}}}\)
\(\displaystyle{\ln{{\left({2.2}\right)}}}={12}{x}{\ln{{\left({1.0038}\right)}}}\)
\(\displaystyle{x}={\frac{{{\ln{{2.2}}}}}{{{12}{\ln{{1.0038}}}}}}\)
\(\displaystyle{x}={17.3236}\).
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