Use the properties of logarithms to evaluate each of the following expressions. (a) \( \log _{12} 3+2 \log _{12} 2=\square \) (b) \( { }^{11} e^{7}-\ln e^{7}=\square \)

(a) We start with \[ \log_{12}3+2\log_{12}2. \] Using the power rule of logarithms, we can write \[ 2\log_{12}2=\log_{12}2^2=\log_{12}4. \] Then, by the product rule, \[ \log_{12}3+\log_{12}4=\log_{12}(3\cdot4)=\log_{12}12. \] Since the logarithm of the base is 1, \[ \log_{12}12=1. \] Thus, the answer for part (a) is \(1\). --- (b) We need to evaluate \[ \log_{11}e^7-\ln e^7. \] First, rewrite the first term using the change-of-base formula: \[ \log_{11}e^7=\frac{\ln(e^7)}{\ln 11}. \] Since \(\ln(e^7)=7\), this becomes \[ \log_{11}e^7=\frac{7}{\ln 11}. \] Also, we know that \[ \ln e^7=7. \] Thus, the expression becomes \[ \frac{7}{\ln 11}-7. \] Factoring out \(7\) gives \[ 7\left(\frac{1}{\ln 11}-1\right). \] So, the answer for part (b) is \(7\left(\frac{1}{\ln 11}-1\right)\).