The value of {log _3}4 times {log _4}5 times | vedclass

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(B) Using the change of base formula,${\log _a}b = \frac{{\log b}}{{\log a}}$.Applying this to the expression:${\log _3}4 	imes {\log _4}5 	imes {\l

Vedclass · Accepted Answer

$2$

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(B) Using the change of base formula,${\log _a}b = \frac{{\log b}}{{\log a}}$.Applying this to the expression:${\log _3}4 	imes {\log _4}5 	imes {\log _5}6 	imes {\log _6}7 	imes {\log _7}8 	imes {\log _8}9 = \frac{{\log 4}}{{\log 3}} 	imes \frac{{\log 5}}{{\log 4}} 	imes \frac{{\log 6}}{{\log 5}} 	imes \frac{{\log 7}}{{\log 6}} 	imes \frac{{\log 8}}{{\log 7}} 	imes \frac{{\log 9}}{{\log 8}}$Canceling the common terms in the numerator and denominator:$= \frac{{\log 9}}{{\log 3}}$$= {\log _3}9$$= {\log _3}({3^2})$$= 2{\log _3}3$$= 2(1) = 2$.

The value of ${\log _3}4 \times {\log _4}5 \times {\log _5}6 \times {\log _6}7 \times {\log _7}8 \times {\log _8}9$ is:

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