The value of log _{3}4 cdot log _{4}5 cdot lo | vedclass

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(B) Using the change of base formula $\log _{a}b = \frac{\log b}{\log a}$:$\log _{3}4 \cdot \log _{4}5 \cdot \log _{5}6 \cdot \log _{6}7 \cdot \log _{

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$2$

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(B) Using the change of base formula $\log _{a}b = \frac{\log b}{\log a}$:$\log _{3}4 \cdot \log _{4}5 \cdot \log _{5}6 \cdot \log _{6}7 \cdot \log _{7}8 \cdot \log _{8}9$$= \frac{\log 4}{\log 3} \cdot \frac{\log 5}{\log 4} \cdot \frac{\log 6}{\log 5} \cdot \frac{\log 7}{\log 6} \cdot \frac{\log 8}{\log 7} \cdot \frac{\log 9}{\log 8}$$= \frac{\log 9}{\log 3}$$= \log _{3}9 = \log _{3}(3^{2}) = 2 \cdot \log _{3}3 = 2 \cdot 1 = 2$.

The value of $\log _{3}4 \cdot \log _{4}5 \cdot \log _{5}6 \cdot \log _{6}7 \cdot \log _{7}8 \cdot \log _{8}9$ is

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