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Thursday, December 31, 2015

2015/118) Without using logs and calculator find the number of digits in 2^{100}

2^{10} = 1024 > 1000 = 10^3
so 2^{100} > 10^{30}
further
2^{10} < 1025 < 10^3 * \frac{41}{40}=  10^3 *( 1+ \frac{1}{40})
so 2^{100} < 10^{30} * ( 1+ \frac{1}{40})^{10} < 10^{30} * ( 1+ \frac{1}{40})^{40}<  10^{30} * e <  10^{30} * 3
so 10^{30} < 2^{100} < 3 * 10^{30}
hence 31 digits

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