Simplify the following using the logarithm laws. \( \begin{array}{ll}\text { a. } \log _{10}(5)+\log _{10}(2) & \text { b. } \log _{2}(80)-\log _{2}(5) \\ \text { c. } \log _{3}\left(2 a^{4}\right)+2 \log _{3}\left(\sqrt{\frac{a}{2}}\right) & \text { d. } \frac{\log _{a}\left(\frac{1}{4}\right)}{\log _{a}(2)}\end{array} \)

Sure! Let's simplify each of the logarithmic expressions using the laws of logarithms. a. \( \log_{10}(5) + \log_{10}(2) = \log_{10}(5 \cdot 2) = \log_{10}(10) = 1 \) b. \( \log_{2}(80) - \log_{2}(5) = \log_{2}\left(\frac{80}{5}\right) = \log_{2}(16) = 4 \) c. \( \log_{3}(2 a^{4}) + 2 \log_{3}\left(\sqrt{\frac{a}{2}}\right) = \log_{3}(2 a^{4}) + \log_{3}\left(\frac{a}{2}\right) = \log_{3}\left(2 a^{4} \cdot \frac{a}{2}\right) = \log_{3}(a^5) = 5 \log_{3}(a) \) d. \( \frac{\log_{a}\left(\frac{1}{4}\right)}{\log_{a}(2)} = \frac{\log_{a}(2^{-2})}{\log_{a}(2)} = \frac{-2 \log_{a}(2)}{\log_{a}(2)} = -2 \) There you go! Each expression has been simplified using logarithm laws.