Question

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

Equations
Solve the equation: $$\displaystyle{{\log}_{{{3}}}{\left({2}{x}+{1}\right)}}={{\log}_{{{3}}}{\left({5}{x}\right)}}+{2}$$.

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 $$\displaystyle{x}={\frac{{{1}}}{{{43}}}}$$