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\( \left|\begin{array}{cc}\log _{a} b & 1 \\ 1 & \log _{b} a\end{array}\right| \) |
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Answer» we know: \( \left |\begin{array}{cc} a & b \\ c & d \end{array} \right| =ad-bc \) and also we know \( \log_a b \times \log_b a=1 \) \( \left |\begin{array}{cc} \log_a b & 1 \\ 1 & \log_b a \end{array} \right| =(\log_a b \times \log_b a) -(1 \times 1)=1-1=0 \) |
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