Consider expression [tex]\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).[/tex]
1. Use property
[tex]\log_a\dfrac{b}{c}=\log_ab-\log_ac.[/tex]
Then
[tex]\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.[/tex]
2. Use property
[tex]\log_abc=\log_ab+\log_ac.[/tex]
Then
[tex]\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2.[/tex]
3. Use property
[tex]\log_ab^k=k\log_ab.[/tex]
Then
[tex]\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+\log_{\frac{1}{2}}x^2-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2.[/tex]
4. Use property
[tex]\log_{a^k}b=\dfrac{1}{k}\log_ab.[/tex]
Then
[tex]\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{\frac{1}{2}}2=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x-\log_{2^{-1}}2=\\ \\=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+\log_22=\log_{\frac{1}{2}}3+2\log_{\frac{1}{2}}x+1.[/tex]
Answer: correct option is B.
