Solve log_{3}(2x-1)-1/4log_{3}(x+1)

$$\log _ { 3 } ( 2 x - 1 ) - \frac { 1 } { 4 } \log _ { 3 } ( x + 1 )$$

Factor

$$\frac{-\ln(x+1)+4\ln(2x-1)}{4\ln(3)}$$

Solution Steps

Factor out \frac{1}{4}.

$$\frac{4\log_{3}\left(2x-1\right)-\log_{3}\left(x+1\right)}{4}$$

Consider 4\ln(2x-1)\times \frac{1}{\ln(3)}-\ln(x+1)\times \frac{1}{\ln(3)}. Factor out \frac{1}{\ln(3)}.

$$\frac{4\ln(2x-1)-\ln(x+1)}{\ln(3)}$$

Rewrite the complete factored expression.

$$\frac{4\ln(2x-1)-\ln(x+1)}{4\ln(3)}$$

Evaluate

$$-\frac{\log_{3}\left(x+1\right)}{4}+\log_{3}\left(2x-1\right)$$