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

Latest revision as of 17:24, 18 October 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

Easily the best solution

Video Solution by OmegaLearn

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

~ pi_is_3.14

2015 AMC 12B (Problems • Answer Key • Resources)
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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 5: / Line 5: @@
 ==Solution 1==
-\begin{align*}
+<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>
-(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==

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