Problem R-8 - solution

SOLUTION TO PROBLEM R-8

Using the change of basis formula one has that

$\begin{displaymath}\log _{a_1}a_2 \cdot \log _{a_2}a_3 \cdot \log _{a_3}a_4 \dots \log _{a_{N-1}}a_N \log _{a_N}a_1 = \end{displaymath}$

$\begin{displaymath}= \frac{\ln a_2}{\ln a_1} \cdot \frac{\ln a_3}{\ln a_2} \cdot... ...s \frac{\ln a_N}{\ln a_{N-1}}\cdot \frac{\ln a_1}{\ln a_N} = 1 \end{displaymath}$

Questions and/or comments should be directed to Judy Downey or Griff Elder

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Last modified: Sat Mar 9 16:31:22 CST 2002