2015 AMC 12B Problems/Problem 8

Revision as of 17:21, 18 October 2024 by Lawofcosine (talk | contribs) (Solution 1)

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

\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 $\log_5 2015$ 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

See Also

2015 AMC 12B (ProblemsAnswer KeyResources)
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

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