## Precalculus: Concepts Through Functions, A Unit Circle Approach to Trigonometry (3rd Edition)

$2 \ln(2x+3)-2 \ln (x-2)-2 \ln (x-1)$
Use the log rule $\log_a{(mn)}=\log_a{m} + \log_a{n}$ We obtain: $\ln[\dfrac{2x+3}{(x-2)(x-1)}]^2 =2[ \ln(2x+3)-\ln (x-1)(x-2)]$ Now, use $\log_a(mn)=\log_a{m}+\log_a{n}$ to obtain: $2[ \ln(2x+3)-[\ln (x-1)+\ln (x-2)]]=2[ \ln(2x+3)-\ln (x-1)-\ln (x-2)] \\=2 \ln(2x+3)-2 \ln (x-2)-2 \ln (x-1)$