Difference between revisions of "2015 AMC 12B Problems/Problem 8"
Pi over two (talk | contribs) (→Solution) |
|||
| Line 4: | Line 4: | ||
<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> | <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} | <math>(625^{\log_5 2015})^\frac{1}{4} | ||
= ((5^4)^{\log_5 2015})^\frac{1}{4} | = ((5^4)^{\log_5 2015})^\frac{1}{4} | ||
| Line 12: | Line 12: | ||
= (2015^4)^\frac{1}{4} | = (2015^4)^\frac{1}{4} | ||
= \boxed{\textbf{(D)}\; 2015}</math> | = \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 = 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 00:22, 11 January 2018
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
Solution 2
We can rewrite 
as as 
. Thus, 
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. 
