The value of frac{log_{3} 5 times log_{25} 27 | vedclass

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Question

(D) Using the change of base formula $\log_{a} b = \frac{\log b}{\log a}$:Numerator = $\log_{3} 5 	imes \log_{25} 27 	imes \log_{49} 7$$= \frac{\log

Vedclass · Accepted Answer

$3$

Vedclass · Answer

(D) Using the change of base formula $\log_{a} b = \frac{\log b}{\log a}$:Numerator = $\log_{3} 5 	imes \log_{25} 27 	imes \log_{49} 7$$= \frac{\log 5}{\log 3} 	imes \frac{\log 27}{\log 25} 	imes \frac{\log 7}{\log 49}$$= \frac{\log 5}{\log 3} 	imes \frac{3 \log 3}{2 \log 5} 	imes \frac{\log 7}{2 \log 7}$$= \frac{1}{1} 	imes \frac{3}{2} 	imes \frac{1}{2} = \frac{3}{4}$Denominator = $\log_{81} 3 = \log_{3^4} 3 = \frac{1}{4} \log_{3} 3 = \frac{1}{4}$Value = $\frac{3/4}{1/4} = 3$

The value of $\frac{\log_{3} 5 \times \log_{25} 27 \times \log_{49} 7}{\log_{81} 3}$ is

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