Which of the following are equivalent to $$ \log _{b}(2) $$ ? $$ \square \log _{b}(12)-\log _{b}(6) $$ $$ \square \frac{1}{2} \log _{b}(4) $$ $$ \square \frac{1}{3} \log _{b}(8) $$ $$ \square-\log _{b}\left(\frac{1}{2}\right) $$ $$ \square \log _{b}\left(\frac{1}{10}\right)+\log _{b}(20) $$

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UpStudy AI · Accepted Answer

1. For $$\log_{b}(12) - \log_{b}(6)$$, use the quotient rule:
   $$
   \log_{b}(12) - \log_{b}(6) = \log_{b}\left(\frac{12}{6}\right) = \log_{b}(2)
   $$

2. For $$\frac{1}{2}\log_{b}(4)$$, note that:
   $$
   \log_{b}(4) = \log_{b}(2^2) = 2\log_{b}(2)
   $$
   Thus,
   $$
   \frac{1}{2}\log_{b}(4) = \frac{1}{2}\cdot 2\log_{b}(2) = \log_{b}(2)
   $$

3. For $$\frac{1}{3}\log_{b}(8)$$, use the fact:
   $$
   \log_{b}(8) = \log_{b}(2^3) = 3\log_{b}(2)
   $$
   So,
   $$
   \frac{1}{3}\log_{b}(8) = \frac{1}{3}\cdot 3\log_{b}(2) = \log_{b}(2)
   $$

4. For $$-\log_{b}\left(\frac{1}{2}\right)$$, recall that:
   $$
   \log_{b}\left(\frac{1}{2}\right) = \log_{b}(2^{-1}) = -\log_{b}(2)
   $$
   Therefore,
   $$
   -\log_{b}\left(\frac{1}{2}\right) = -\left(-\log_{b}(2)\right) = \log_{b}(2)
   $$

5. For $$\log_{b}\left(\frac{1}{10}\right) + \log_{b}(20)$$, apply the product rule:
   $$
   \log_{b}\left(\frac{1}{10}\right) + \log_{b}(20) = \log_{b}\left(\frac{1}{10} \cdot 20\right) = \log_{b}(2)
   $$

Thus, all the given expressions are equivalent to $$\log_{b}(2)$$.
All the given expressions are equivalent to $$ \log_{b}(2) $$.

Which of the following are equivalent to \( \log _{b}(2) \) ? \( \square \log _{b}(12)-\log _{b}(6) \) \( \square \frac{1}{2} \log _{b}(4) \) \( \square \frac{1}{3} \log _{b}(8) \) \( \square-\log _{b}\left(\frac{1}{2}\right) \) \( \square \log _{b}\left(\frac{1}{10}\right)+\log _{b}(20) \)

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