8. a) \( \log _{2}\left(2^{m}\right) \) \( \begin{array}{lllll}\text { b) } \log _{5}\left(\frac{1}{5^{m}}\right) & \text { d) } \log _{4}\left(\frac{1}{4^{\frac{1}{n}}}\right) & \text { g) } \log _{b}\left(\frac{1}{b}\right) & \text { j) } \log _{a}\left(a^{n}\right) & \text { m) } \log _{a}\left(a^{\frac{1}{n}}\right) \\ \text { c) } \log _{3}\left(3^{\frac{1}{n}}\right) & \text { f) } \log _{b}\left(b^{2}\right) & \text { k) } \log _{c}(\sqrt{c}) & \text { n) } \log _{a}\left(a^{\frac{m}{n}}\right)\end{array} \)

**a)** Using the logarithm property \(\log_b(b^x) = x\), we have \[ \log_{2}(2^{m}) = m. \] **b)** Rewrite \(\frac{1}{5^{m}}\) as \(5^{-m}\). Then, \[ \log_{5}\left(\frac{1}{5^{m}}\right)=\log_{5}\left(5^{-m}\right)=-m. \] **c)** Express the argument as a power of 3: \[ \log_{3}\left(3^{\frac{1}{n}}\right)=\frac{1}{n}. \] **d)** Rewrite \(\frac{1}{4^{\frac{1}{n}}}\) as \(4^{-\frac{1}{n}}\). Then, \[ \log_{4}\left(\frac{1}{4^{\frac{1}{n}}}\right)=\log_{4}\left(4^{-\frac{1}{n}}\right)=-\frac{1}{n}. \] **f)** Using the logarithm property directly, \[ \log_{b}\left(b^2\right)=2. \] **g)** Express the argument as a power of \(b\): \[ \log_{b}\left(\frac{1}{b}\right)=\log_{b}\left(b^{-1}\right)=-1. \] **j)** Using the property \(\log_a(a^x)=x\), \[ \log_{a}\left(a^n\right)=n. \] **k)** Rewrite \(\sqrt{c}\) as \(c^{\frac{1}{2}}\). Then, \[ \log_{c}\left(\sqrt{c}\right)=\log_{c}\left(c^{\frac{1}{2}}\right)=\frac{1}{2}. \] **m)** Similarly, \[ \log_{a}\left(a^{\frac{1}{n}}\right)=\frac{1}{n}. \] **n)** Expressing the argument as a power of \(a\), \[ \log_{a}\left(a^{\frac{m}{n}}\right)=\frac{m}{n}. \]