Difference between revisions of "2015 AMC 12B Problems/Problem 8"
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==Solution== | ==Solution== | ||
| − | + | <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> | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2015|ab=B|num-a=9|num-b=7}} | {{AMC12 box|year=2015|ab=B|num-a=9|num-b=7}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 23:29, 3 March 2015
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
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 | |
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