2010 AMC 8 Problems/Problem 24

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Problem

What is the correct ordering of the three numbers, $10^8$, $5^{12}$, and $2^{24}$?

$\textbf{(A)}\ 2^2^4<10^8<5^1^2$ (Error compiling LaTeX. Unknown error_msg) $\textbf{(B)}\ 2^2^4<5^1^2<10^8$ (Error compiling LaTeX. Unknown error_msg) $\textbf{(C)}\ 5^1^2<2^2^4<10^8$ (Error compiling LaTeX. Unknown error_msg) $\textbf{(D)}\ 10^8<5^1^2<2^2^4$ (Error compiling LaTeX. Unknown error_msg) $\textbf{(E)}\ 10^8<2^2^4<5^1^2$ (Error compiling LaTeX. Unknown error_msg)

Solution

Since all of the exponents are multiples of four, we can simplify the problem by taking the fourth root of each number. Evaluating we get $10^2=100$, $5^3=125$, and $2^6=64$. Since $64<100<125$, it follows that $\boxed{\textbf{(A)}\ 2^2^4<10^8<5^1^2 }$ (Error compiling LaTeX. Unknown error_msg) is the correct answer.

See Also

2011 AMC 8 (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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All AJHSME/AMC 8 Problems and Solutions