Which of the following are equivalent to \( \log _{b}(2) \) ? \[ \log _{b}\left(\frac{1}{10}\right)+\log _{b}(20) \] \( \square \frac{1}{3} \log _{b}(8) \) \( \square \log _{b}(16)-\log _{b}(8) \) \( \square \frac{1}{2} \log _{b}(4) \) \( \square-\log _{b}\left(\frac{1}{2}\right) \)

To determine which expressions are equivalent to \( \log_{b}(2) \), let's analyze each one. 1. **\( \log_{b}\left(\frac{1}{10}\right) + \log_{b}(20) \):** Using the property \( \log_{b}(x) + \log_{b}(y) = \log_{b}(xy) \), we can combine these: \[ \log_{b}\left(\frac{1}{10} \cdot 20\right) = \log_{b}(2) \] Thus, this expression is equivalent to \( \log_{b}(2) \). 2. **\( \frac{1}{3} \log_{b}(8) \):** We know that \( \log_{b}(8) = \log_{b}(2^3) = 3 \log_{b}(2) \), so: \[ \frac{1}{3} \log_{b}(8) = \frac{1}{3} (3 \log_{b}(2)) = \log_{b}(2) \] This expression is equivalent to \( \log_{b}(2) \). 3. **\( \log_{b}(16) - \log_{b}(8) \):** Using the property \( \log_{b}(x) - \log_{b}(y) = \log_{b}\left(\frac{x}{y}\right) \): \[ \log_{b}\left(\frac{16}{8}\right) = \log_{b}(2) \] This expression is equivalent to \( \log_{b}(2) \). 4. **\( \frac{1}{2} \log_{b}(4) \):** Since \( \log_{b}(4) = \log_{b}(2^2) = 2 \log_{b}(2) \), we have: \[ \frac{1}{2} \log_{b}(4) = \frac{1}{2} (2 \log_{b}(2)) = \log_{b}(2) \] So this expression is also equivalent to \( \log_{b}(2) \). 5. **\( -\log_{b}\left(\frac{1}{2}\right) \):** We know that \( \log_{b}\left(\frac{1}{2}\right) = \log_{b}(2^{-1}) = -\log_{b}(2) \), therefore: \[ -\log_{b}\left(\frac{1}{2}\right) = -(-\log_{b}(2)) = \log_{b}(2) \] Hence, this expression is equivalent to \( \log_{b}(2) \). In conclusion, all five options provided are equivalent to \( \log_{b}(2) \): - \( \log_{b}\left(\frac{1}{10}\right) + \log_{b}(20) \) is equivalent. - \( \frac{1}{3} \log_{b}(8) \) is equivalent. - \( \log_{b}(16) - \log_{b}(8) \) is equivalent. - \( \frac{1}{2} \log_{b}(4) \) is equivalent. - \( -\log_{b}\left(\frac{1}{2}\right) \) is equivalent.