If log 2 = 0.3010 and log 3 = 0.4771, then th | vedclass

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(C) Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$.We need to find the value of $\log _{5} 512$.First,express $512$ as a power of $2$: $512 = 2^9$.Usi

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$3.876$

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(C) Given: $\log 2 = 0.3010$ and $\log 3 = 0.4771$.We need to find the value of $\log _{5} 512$.First,express $512$ as a power of $2$: $512 = 2^9$.Using the change of base formula,$\log _{b} a = \frac{\log _{10} a}{\log _{10} b}$.Therefore,$\log _{5} 512 = \log _{5} (2^9) = 9 \log _{5} 2$.This can be written as $9 	imes \frac{\log _{10} 2}{\log _{10} 5}$.Since $\log _{10} 5 = \log _{10} (10/2) = \log _{10} 10 - \log _{10} 2 = 1 - 0.3010 = 0.6990$.Substituting the values: $\log _{5} 512 = \frac{9 	imes 0.3010}{0.6990} = \frac{2.709}{0.699} = 3.8755... \approx 3.876$.

If $\log 2 = 0.3010$ and $\log 3 = 0.4771,$ then the value of $\log _{5} 512$ is

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