\[ \text { ClufN } y=\log _{g} 8 \] (a) \( \log _{q} 64 \) (b) \( \log _{2} 2 \) (c) \( \log _{q} \frac{16}{q} \)

To solve the logs in terms of the given \( y = \log_{g} 8 \), we can express each term using properties of logarithms. First, remember that \( \log_{b}(a^n) = n \cdot \log_{b}(a) \). For \( \log_{q} 64 \), we note that \( 64 = 8^2 \), so: \[ \log_{q} 64 = \log_{q} (8^2) = 2 \cdot \log_{q} 8 \] Next, \( \log_{2} 2 \) is simply equal to 1 since any logarithm of a number to its own base is 1. For \( \log_{q} \frac{16}{q} \), we can apply the properties of logarithms: \[ \log_{q} \frac{16}{q} = \log_{q} 16 - \log_{q} q = \log_{q} 16 - 1 \] Now we can express \( \log_{q} 16 \) in relation to \( y \). Noticing that \( 16 = 2^4 \) and \( 8 = 2^3 \), we adjust \( y \) for conformity when we convert bases. So, based on these expressions: 1. \( \log_{q} 64 = 2 \cdot \log_{q} 8 \) 2. \( \log_{2} 2 = 1 \) 3. \( \log_{q} \frac{16}{q} = \log_{q} 16 - 1 \) Hope these transformations make your calculations as smooth as possible!