Difference between revisions of "2015 AMC 12B Problems/Problem 8"

Revision as of 00:13, 27 March 2024

Problem

What is the value of $(625^{\log_5 2015})^{\frac{1}{4}}$ ?

$\textbf{(A)}\; 5 \qquad\textbf{(B)}\; \sqrt[4]{2015} \qquad\textbf{(C)}\; 625 \qquad\textbf{(D)}\; 2015 \qquad\textbf{(E)}\; \sqrt[4]{5^{2015}}$

Solution 1

$(625^{\log_5 2015})^\frac{1}{4} = ((5^4)^{\log_5 2015})^\frac{1}{4} = (5^{4 \cdot \log_5 2015})^\frac{1}{4} = (5^{\log_5 2015 \cdot 4})^\frac{1}{4} = ((5^{\log_5 2015})^4)^\frac{1}{4} = (2015^4)^\frac{1}{4} = \boxed{\textbf{(D)}\; 2015}$

Solution 2

We can rewrite $\log_5 2015$ as as $5^x = 2015$ . Thus, $625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.$

Solution 3

$(625^{\log_5 2015})^{\frac{1}{4}} = (625^{\frac{1}{4}})^{\log_5 2015} = 5^{\log_5 2015} = \boxed{\textbf{(D)} 2015}$

~ cxsmi

Solution 4 (Last resort)

We note that the year number is just $2015$ , so just guess $\boxed{\textbf{(D)} 2015}.

~xHypotenuse

Video Solution by OmegaLearn

https://youtu.be/RdIIEhsbZKw?t=738

~ pi_is_3.14

2015 AMC 12B (Problems • Answer Key • Resources)
Preceded by Problem 7	Followed by Problem 9
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All AMC 12 Problems and Solutions

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.

@@ Line 20: / Line 20: @@
 ~ cxsmi
+==Solution 4 (Last resort)==
+We note that the year number is just <math>2015</math>, so just guess $\boxed{\textbf{(D)} 2015}.
+~xHypotenuse
 == Video Solution by OmegaLearn ==

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