If log_{7} 2 = m,then log_{49} 28 is equal to | vedclass

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(B) Given $\log_{7} 2 = m$.We need to evaluate $\log_{49} 28$.Using the change of base formula,$\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$,we have:$\

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$\frac{1 + 2m}{2}$

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(B) Given $\log_{7} 2 = m$.We need to evaluate $\log_{49} 28$.Using the change of base formula,$\log_{a} b = \frac{\log_{c} b}{\log_{c} a}$,we have:$\log_{49} 28 = \frac{\log_{7} 28}{\log_{7} 49}$Since $28 = 7 	imes 4 = 7 	imes 2^2$ and $49 = 7^2$,we get:$\log_{49} 28 = \frac{\log_{7} (7 	imes 2^2)}{\log_{7} (7^2)}$Using the properties $\log(xy) = \log x + \log y$ and $\log(x^n) = n \log x$:$\log_{49} 28 = \frac{\log_{7} 7 + \log_{7} 2^2}{2 \log_{7} 7}$$\log_{49} 28 = \frac{1 + 2 \log_{7} 2}{2(1)}$Substituting $\log_{7} 2 = m$:$\log_{49} 28 = \frac{1 + 2m}{2}$

If $\log_{7} 2 = m$,then $\log_{49} 28$ is equal to

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