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given logbase10 2=0.30103, find number of digits in 2000^2000.?
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- JBLv 76 years agoFavorite Answer
The length in digits of a large number is essentially the log base 10 of the number.
log[10] (2000^2000) = 2000 log[10] (2 * 1000) =
2000 * (log[10] 1000 + log[10] 2) =
2000 * ( 3 + 0.30103) ≈ 6602 digits.
(the exact answer is 6603 digits).
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