Solve 2/3ln(x+2)^2-1/2[ln(x^2+3x+2)^2]

$$\frac { 2 } { 3 } \ln ( x + 2 ) ^ { 2 } - \frac { 1 } { 2 } [ \ln ( x ^ { 2 } + 3 x + 2 ) ^ { 2 } ]$$

Factor

$$\frac{-3\ln(\left(x+1\right)^{2}\left(x+2\right)^{2})+4\ln(\left(x+2\right)^{2})}{6}$$

Solution Steps

Factor out \frac{1}{6}.

$$\frac{4\log_{e}\left(\left(x+2\right)^{2}\right)-3\log_{e}\left(\left(x^{2}+3x+2\right)^{2}\right)}{6}$$

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

$$\frac{4\ln(\left(x+2\right)^{2})-3\ln(\left(x^{2}+3x+2\right)^{2})}{\ln(e)}$$

Rewrite the complete factored expression.

$$\frac{4\ln(\left(x+2\right)^{2})-3\ln(\left(x^{2}+3x+2\right)^{2})}{6\ln(e)}$$

Evaluate

$$\frac{-3\ln(\left(\left(x+1\right)\left(x+2\right)\right)^{2})+4\ln(\left(x+2\right)^{2})}{6}$$