If log_{4}5 = a and log_{5}6 = b,then log_{3} | vedclass

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(D) Given: $\log_{4}5 = a$ and $\log_{5}6 = b$.Using the change of base formula,$ab = \log_{4}5 	imes \log_{5}6 = \log_{4}6$.Since $\log_{4}6 = \frac

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$\frac{1}{2ab - 1}$

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(D) Given: $\log_{4}5 = a$ and $\log_{5}6 = b$.Using the change of base formula,$ab = \log_{4}5 	imes \log_{5}6 = \log_{4}6$.Since $\log_{4}6 = \frac{\log_{2}6}{\log_{2}4} = \frac{\log_{2}(2 	imes 3)}{2} = \frac{1 + \log_{2}3}{2}$.So,$ab = \frac{1 + \log_{2}3}{2}$.Multiplying by $2$,we get $2ab = 1 + \log_{2}3$,which implies $\log_{2}3 = 2ab - 1$.Therefore,$\log_{3}2 = \frac{1}{\log_{2}3} = \frac{1}{2ab - 1}$.

If $\log_{4}5 = a$ and $\log_{5}6 = b$,then $\log_{3}2$ is equal to

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