Difference between revisions of "2015 AMC 12B Problems/Problem 8"
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==Solution 1== | ==Solution 1== | ||
| − | <math>(625^{\log_5 2015})^\frac{1}{4} | + | <math>(625^{\log_5 2015})^\frac{1}{4}=((5^4)^{\log_5 2015})^\frac{1}{4}=(5^{4 \cdot \log_5 2015})^\frac{1}{4}=(5^{\log_5 2015 \cdot 4})^\frac{1}{4}=((5^{\log_5 2015})^4)^\frac{1}{4}=(2015^4)^\frac{1}{4}=\boxed{\textbf{(D)}\; 2015}</math> |
| − | = ((5^4)^{\log_5 2015})^\frac{1}{4} | ||
| − | = (5^{4 \cdot \log_5 2015})^\frac{1}{4} | ||
| − | = (5^{\log_5 2015 \cdot 4})^\frac{1}{4} | ||
| − | = ((5^{\log_5 2015})^4)^\frac{1}{4} | ||
| − | = (2015^4)^\frac{1}{4} | ||
| − | = \boxed{\textbf{(D)}\; 2015}</math> | ||
==Solution 2== | ==Solution 2== | ||
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==Solution 3== | ==Solution 3== | ||
| − | <math>(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{\textbf{(D)} 2015}</math> | + | <math>(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{\textbf{(D)}~2015}</math> |
| − | ~ cxsmi | + | ~ [https://artofproblemsolving.com/wiki/index.php/User:Cxsmi cxsmi] |
| + | |||
| + | ==Solution 4 (Last resort)== | ||
| + | We note that the year number is just <math>2015</math>, so just guess <math>\boxed{\textbf{(D)} 2015}</math>. | ||
| + | |||
| + | ~xHypotenuse | ||
| + | |||
| + | Easily the best solution | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
Latest revision as of 17:24, 18 October 2024
Contents
Problem
What is the value of 
?
![$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$](http://latex.artofproblemsolving.com/9/1/2/9123c6861615bbff4b77a8e0d9774474701447a6.png)
Solution 1
Solution 2
We can rewrite 
as as 
. Thus, 
Solution 3
~ cxsmi
Solution 4 (Last resort)
We note that the year number is just 
, so just guess 
.
~xHypotenuse
Easily the best solution
Video Solution by OmegaLearn
https://youtu.be/RdIIEhsbZKw?t=738
~ pi_is_3.14
See Also
| 2015 AMC 12B (Problems • Answer Key • Resources) | |
| Preceded by Problem 7 |
Followed by Problem 9 |
| 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 AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
