Difference between revisions of "2015 AMC 12B Problems/Problem 8"
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==Solution 1== | ==Solution 1== | ||
\begin{align*} | \begin{align*} | ||
| − | (625^{\log_5 2015})^\frac{1}{4} &= ((5^4)^{\log_5 2015})^\frac{1}{4}\\ &= (5^{4 \cdot \log_5 2015})^\frac{1}{4}\\ | + | (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 \cdot 4})^\frac{1}{4}\\ | ||
&= ((5^{\log_5 2015})^4)^\frac{1}{4}\\ | &= ((5^{\log_5 2015})^4)^\frac{1}{4}\\ | ||
Revision as of 17:21, 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
\begin{align*} (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}\\
\end{align*}
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. 
