Difference between revisions of "1993 AHSME Problems/Problem 11"
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| + | Taking successive exponentials <math>\log_2(\log_2(x)) = 2^2 = 4</math> and <math>\log_2(x) = 2^4=16</math> and <math>x = 2^{16}</math>. Now <math>2^{10} = 1024 \approx 10^3</math> and <math>2^6 = 64</math> so we can approximate <math>2^{16} \approx 64000</math> which has 5 digits. In general, <math>2^n</math> has approximately <math>n/3</math> digits. | ||
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Revision as of 21:10, 27 May 2021
Problem
If 
, then how many digits are in the base-ten representation for x?
Solution
Taking successive exponentials 
and 
and 
. Now 
and 
so we can approximate 
which has 5 digits. In general, 
has approximately 
digits.
See also
| 1993 AHSME (Problems • Answer Key • Resources) | ||
| Preceded by Problem 10 |
Followed by Problem 12 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 • 26 • 27 • 28 • 29 • 30 | ||
| All AHSME Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
