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

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(Solution 2)
 
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==Solution 2==
 
==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 = \boxed{2015}.</math>
 
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 = \boxed{2015}.</math>
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== Video Solution ==
 +
https://youtu.be/RdIIEhsbZKw?t=738
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 +
~ pi_is_3.14
  
 
==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}}

Latest revision as of 03:29, 21 January 2021

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, $625^{x \cdot \frac{1}{4}} = 5^x = \boxed{2015}.$

Video Solution

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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