The number of digits in 20^{301} (given, log | vedclass

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Question

(C) Let $y = 20^{301}$.To find the number of digits in $y$, we calculate $\lfloor \log_{10} y \rfloor + 1$.$\log_{10} y = \log_{10} (20^{301}) = 301 \

Vedclass · Accepted Answer

$392$

Vedclass · Answer

(C) Let $y = 20^{301}$.To find the number of digits in $y$, we calculate $\lfloor \log_{10} y floor + 1$.$\log_{10} y = \log_{10} (20^{301}) = 301 	imes \log_{10} (2 	imes 10)$.Using the property $\log(ab) = \log a + \log b$, we get:$\log_{10} y = 301 	imes (\log_{10} 2 + \log_{10} 10) = 301 	imes (0.3010 + 1) = 301 	imes 1.3010$.$301 	imes 1.3010 = 391.601$.The number of digits is $\lfloor 391.601 floor + 1 = 391 + 1 = 392$.

The number of digits in $20^{301}$ (given, $\log _{10} 2=0.3010$) is:

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