Difference between revisions of "2010 AMC 8 Problems/Problem 24"
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What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>? | What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>? | ||
| − | <math> \textbf{(A)}\ 2^ | + | <math> \textbf{(A)}\ 2^{24}<10^8<5^{12}</math> |
| − | <math> \textbf{(B)}\ 2^ | + | <math> \textbf{(B)}\ 2^{24}<5^{12}<10^8 </math> |
| − | <math> \textbf{(C)}\ 5^1^2<2^ | + | <math> \textbf{(C)}\ 5^{1^2}<2^{24}<10^8 </math> |
| − | <math> \textbf{(D)}\ 10^8<5^ | + | <math> \textbf{(D)}\ 10^8<5^{12}<2^{24}</math> |
| − | <math> \textbf{(E)}\ 10^8<2^ | + | <math> \textbf{(E)}\ 10^8<2^{24}<5^{12} </math> |
==Solution== | ==Solution== | ||
Revision as of 17:05, 18 March 2015
Problem
What is the correct ordering of the three numbers, 
, 
, and 
?
Solution
Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get 
, 
, and 
. Since 
, it follows that $\boxed{\textbf{(A)}\ 2^2^4<10^8<5^1^2 }$ (Error compiling LaTeX. Unknown error_msg) is the correct answer.
See Also
| 2010 AMC 8 (Problems • Answer Key • Resources) | ||
| Preceded by Problem 23 |
Followed by Problem 25 | |
| 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 | ||
| All AJHSME/AMC 8 Problems and Solutions | ||
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
