2015 AMC 12B Problems/Problem 8: Difference between revisions
Orion 2010 (talk | contribs) |
|||
| (7 intermediate revisions by one other user not shown) | |||
| Line 5: | Line 5: | ||
==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== | ||
| Line 27: | Line 21: | ||
Easily the best solution | Easily the best solution | ||
(yeah definetly) | |||
== Video Solution by OmegaLearn == | == Video Solution by OmegaLearn == | ||
Latest revision as of 12:36, 31 July 2025
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 (yeah definetly)
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 | |
These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions. Error creating thumbnail: File missing
