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

**a. Simplify \(\log_{10}(5)+\log_{10}(2)\)** Using the logarithm product law: \[ \log_{10}(5) + \log_{10}(2) = \log_{10}(5 \cdot 2) = \log_{10}(10). \] Since \(\log_{10}(10)=1\), the simplified expression is: \[ 1. \] --- **b. Simplify \(\log_{2}(80)-\log_{2}(5)\)** Using the logarithm quotient law: \[ \log_{2}(80) - \log_{2}(5) = \log_{2}\left(\frac{80}{5}\right) = \log_{2}(16). \] Since \(16=2^4\), we have: \[ \log_{2}(16)=4. \] Thus, the simplified expression is: \[ 4. \] --- **c. Simplify \(\log_{3}\left(2a^{4}\right)+2\log_{3}\left(\sqrt{\frac{a}{2}}\right)\)** First, express the square root in exponent form: \[ \sqrt{\frac{a}{2}}=\left(\frac{a}{2}\right)^{\frac{1}{2}}. \] Then, using the power law on the second term: \[ 2\log_{3}\left(\sqrt{\frac{a}{2}}\right)=2\log_{3}\left(\left(\frac{a}{2}\right)^{\frac{1}{2}}\right)=2\cdot\frac{1}{2}\log_{3}\left(\frac{a}{2}\right)=\log_{3}\left(\frac{a}{2}\right). \] Now, combine the logarithms using the product law: \[ \log_{3}\left(2a^{4}\right)+\log_{3}\left(\frac{a}{2}\right)=\log_{3}\left(2a^{4}\cdot\frac{a}{2}\right). \] Simplify inside the logarithm: \[ 2a^{4}\cdot\frac{a}{2}=a^{4}\cdot a=a^{5}. \] Thus, the expression simplifies to: \[ \log_{3}\left(a^{5}\right). \] --- **d. Simplify \(\frac{\log_{a}\left(\frac{1}{4}\right)}{\log_{a}(2)}\)** Using the change of base formula: \[ \frac{\log_{a}\left(\frac{1}{4}\right)}{\log_{a}(2)}=\log_{2}\left(\frac{1}{4}\right). \] Recognize that \(\frac{1}{4}=4^{-1}\), so: \[ \log_{2}\left(4^{-1}\right)=-\log_{2}(4). \] Since \(4=2^2\), it follows that: \[ \log_{2}(4)=2. \] Thus, the expression simplifies to: \[ -2. \]