Difference between revisions of "2015 AMC 12B Problems/Problem 8"
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==Solution 2== | ==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> | 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> | ||
+ | |||
+ | ~ cxsmi | ||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == |
Revision as of 14:06, 10 February 2024
Problem
What is the value of ?
Solution 1
Solution 2
We can rewrite as as
. Thus,
Solution 3
~ cxsmi
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.