Question

Solve the equation: \log_{3}(2x+1)=\log_{3}(5x)+2.

Equations
ANSWERED
asked 2021-02-21
Solve the equation: \(\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}={{\log}_{{{3}}}{\left({5}{x}\right)}}+{2}\).

Answers (1)

2021-02-23
Step 1
Given equation:
\(\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}={{\log}_{{{3}}}{\left({5}{x}\right)}}+{2}\)
Step 2
Now
Apply \(\displaystyle{\log{}}\) rule: \(\displaystyle{{\log}_{{{c}}}{\left({a}\right)}}-{{\log}_{{{c}}}{\left({b}\right)}}={{\log}_{{{c}}}{\left({\frac{{{a}}}{{{b}}}}\right)}}\)
Apply \(\displaystyle{\log{}}\) rule: \(\displaystyle{{\log}_{{{c}}}{\left({a}\right)}}={y}\Rightarrow{a}={c}^{{{y}}}\)
Therefore,
\(\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}={{\log}_{{{3}}}{\left({5}{x}\right)}}+{2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}-{{\log}_{{{3}}}{\left({5}{x}\right)}}={2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({\frac{{{2}{x}+{1}}}{{{5}{x}}}}\right)}}={2}\)
\(\displaystyle{\frac{{{2}{x}+{1}}}{{{5}{x}}}}={3}^{{{2}}}\)
\(\displaystyle{\frac{{{2}{x}+{1}}}{{{5}{x}}}}={9}\)
\(\displaystyle{2}{x}+{1}={9}\times{5}{x}\)
2+1=45x
1=45x-2x
43x=1
\(\displaystyle{x}={\frac{{{1}}}{{{43}}}}\)
To verify the solution, put \(\displaystyle{x}={\frac{{{1}}}{{{43}}}}\) in the given equation:
\(\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}={{\log}_{{{3}}}{\left({5}{x}\right)}}+{2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}-{{\log}_{{{3}}}{\left({5}{x}\right)}}={2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({\frac{{{2}{x}+{1}}}{{{5}{x}}}}\right)}}={2}\)
Put \(\displaystyle{x}={\frac{{{1}}}{{{43}}}}\)
\(\displaystyle{{\log}_{{{3}}}{\left({\frac{{{2}\times{\frac{{{1}}}{{{43}}}}+{1}}}{{{5}\times{\frac{{{1}}}{{{43}}}}}}}\right)}}={2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({\frac{{{45}}}{{{5}}}}\right)}}={2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({9}\right)}}={2}\)
\(\displaystyle{{\log}_{{{3}}}{\left({3}^{{{2}}}\right)}}={2}\)
Apply \(\displaystyle{\log{}}\) rule: \(\displaystyle{{\log}_{{{c}}}{\left({c}^{{{x}}}\right)}}={x}\)
2=2
\(\displaystyle\therefore\) LHS=RHS
Hence,
The required solution is \(\displaystyle{x}={\frac{{{1}}}{{{43}}}}\)
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