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Solve logarithmic equation ln(3x + 1) - ln(5 +x) = ln 2
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start with \[\ln(\frac{3x+1}{x+5})=\ln(2)\] then go right to \[\frac{3x+1}{x+5}=2\]
\[ \large \begin{array}{rcl} \ln(3x + 1) - \ln(5 +x) &=& \ln 2 \\ \ln\left(\frac{3x + 1}{5 +x}\right) &=& \ln 2 \\ e^{\ln\left(\frac{3x + 1}{5 +x}\right)} &=& e^{\ln 2} \\ \frac{3x + 1}{5 +x} &=& 2 \end{array} \]
what @wio said, although since log is a one to one function you do not need to exponentiate \[\ln(a)=\ln(b)\iff a = b\]
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