Log Base 2

What is the value of
\dfrac{log_2 3\cdot log_4 5\cdot log_6 7}{log_4 3\cdot log_6 5\cdot log_8 7}
Source: NCTM Mathematics Teacher, November 2006

SOLUTION
We change all the different bases to base 10
log_2 3=log\,3/log\,2
log_4 5=log\,5/log\,4
log_6 7=log\,7/log\,6
log_4 3=log\,3/log\,4
log_6 5=log\,5/log\,6
log_8 7=log\,7/log\,8

\dfrac{log_2 3\cdot log_4 5\cdot log_6 7}{log_4 3\cdot log_6 5\cdot log_8 7}=\dfrac{log\,3/log\,2\cdot log\,5/log\,4\cdot log\,7/log\,6}{log\,3/log\,4\cdot log\,5/log\,6\cdot log\,7/log\,8}
=\dfrac{1/log\,2}{1/log\,8}
=log\,8/log\,2
=log\,2^3/log\,2
=3\,log\,2/log\,2
=3

Answer: 3

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About mvtrinh

Retired high school math teacher.
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