Difference between revisions of "2015 AMC 12B Problems/Problem 8"
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What is the value of <math>(625^{\log_5 2015})^{\frac{1}{4}}</math> ? | What is the value of <math>(625^{\log_5 2015})^{\frac{1}{4}}</math> ? | ||
− | <math>\textbf{(A)}\; | + | <math>\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}</math> |
− | ==Solution== | + | ==Solution 1== |
+ | <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> | ||
+ | ==Solution 2== | ||
+ | We can rewrite <math>\log_5 2015</math> as as <math>5^x = 2015</math>. Thus, <math>625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.</math> | ||
+ | |||
+ | ==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> | ||
+ | |||
+ | ~ [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 == | ||
+ | https://youtu.be/RdIIEhsbZKw?t=738 | ||
+ | |||
+ | ~ pi_is_3.14 | ||
==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}} |
Latest revision as of 02:14, 30 June 2024
Contents
Problem
What is the value of ?
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.