Difference between revisions of "1993 AHSME Problems/Problem 11"

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== Solution ==
 
== Solution ==
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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 22:10, 27 May 2021

Problem

If $log_2(log_2(log_2(x)))=2$, then how many digits are in the base-ten representation for x?

$\text{(A) } 5\quad \text{(B) } 7\quad \text{(C) } 9\quad \text{(D) } 11\quad \text{(E) } 13$

Solution

Taking successive exponentials $\log_2(\log_2(x)) = 2^2 = 4$ and $\log_2(x) = 2^4=16$ and $x = 2^{16}$. Now $2^{10} = 1024 \approx 10^3$ and $2^6 = 64$ so we can approximate $2^{16} \approx 64000$ which has 5 digits. In general, $2^n$ has approximately $n/3$ digits.

$\fbox{A}$

See also

1993 AHSME (ProblemsAnswer KeyResources)
Preceded by
Problem 10
Followed by
Problem 12
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