If log 2 = 0.30103 and log 3 = 0.4771,then th | vedclass

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(D) Let $x = (648)^{5}$.Taking logarithm on both sides,we get $\log x = 5 \log (648)$.Since $648 = 2^3 	imes 3^4$,we have $\log x = 5 \log (2^3 	ime

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$15$

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(D) Let $x = (648)^{5}$.Taking logarithm on both sides,we get $\log x = 5 \log (648)$.Since $648 = 2^3 	imes 3^4$,we have $\log x = 5 \log (2^3 	imes 3^4)$.Using the properties of logarithms,$\log x = 5 (3 \log 2 + 4 \log 3)$.Substituting the given values,$\log x = 5 (3 	imes 0.30103 + 4 	imes 0.4771)$.$\log x = 5 (0.90309 + 1.9084) = 5 (2.81149) = 14.05745$.The number of digits in a number $x$ is given by $\lfloor \log_{10} x floor + 1$.Here,$\lfloor 14.05745 floor + 1 = 14 + 1 = 15$.Therefore,the number of digits in $(648)^{5}$ is $15$.

If $\log 2 = 0.30103$ and $\log 3 = 0.4771$,then the number of digits in $(648)^{5}$ is

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