2010 AMC 8 Problems/Problem 24: Difference between revisions
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==Solution== | ==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 <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^ | 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 <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer. | ||
==See Also== | ==See Also== | ||
{{AMC8 box|year=2010|num-b=23|num-a=25}} | {{AMC8 box|year=2010|num-b=23|num-a=25}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 16:07, 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 
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 | ||
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