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faltersainse [42]
3 years ago
5

Expand using the properties and rules for logarithms

Mathematics
1 answer:
malfutka [58]3 years ago
7 0

Consider expression \log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right).

1. Use property

\log_a\dfrac{b}{c}=\log_ab-\log_ac.

Then

\log_{\frac{1}{2}}\left(\dfrac{3x^2}{2}\right)=\log_{\frac{1}{2}}3x^2-\log_{\frac{1}{2}}2.

2. Use property

\log_abc=\log_ab+\log_ac.

Then

\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.

3. Use property

\log_ab^k=k\log_ab.

Then

\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.

4. Use property

\log_{a^k}b=\dfrac{1}{k}\log_ab.

Then

\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.

Answer: correct option is B.

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