2010 AMC 8 Problem 24 — Solution | LIVE by Po-Shen Loh

LIVE

English

2010 AMC 8 Problem 24

Attempt Problem 24 of the 2010 AMC 8 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 2010 AMC 8 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

24.

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

2^{24} < 10^8 < 5^{12}

2^{24} < 5^{12} < 10^8

5^{12} < 2^{24} < 10^8

10^8 < 5^{12} < 2^{24}

10^8 < 2^{24} < 5^{12}

Answer: A

Concepts:exponent

Difficulty rating: 1480

Solution:

First, we get $2^{24} = (2^8)(4^8) < (2^8)(5^8) = 10^8.$

Next, we get $10^8 = (2^8)(5^8) =$ $(4^4)(5^8) < (5^4)(5^8) = 5^{12}.$ This means $2^{24} < 10^8 < 5^{12} .$

Thus, the answer is A .

← Problem 23#23

Problem 25#25 →

Problem 24 in Other Years

1985 AMC 8 · 1986 AMC 8 · 1987 AMC 8 · 1988 AMC 8 · 1989 AMC 8 · 1990 AMC 8 · 1991 AMC 8 · 1992 AMC 8 · 1993 AMC 8 · 1994 AMC 8 · 1995 AMC 8 · 1996 AMC 8 · 1997 AMC 8 · 1998 AMC 8 · 1999 AMC 8 · 2000 AMC 8 · 2001 AMC 8 · 2002 AMC 8 · 2003 AMC 8 · 2004 AMC 8 · 2005 AMC 8 · 2006 AMC 8 · 2007 AMC 8 · 2008 AMC 8 · 2009 AMC 8 · 2011 AMC 8 · 2012 AMC 8 · 2013 AMC 8 · 2014 AMC 8 · 2015 AMC 8 · 2016 AMC 8 · 2017 AMC 8 · 2018 AMC 8 · 2019 AMC 8 · 2020 AMC 8 · 2022 AMC 8 · 2023 AMC 8 · 2024 AMC 8 · 2025 AMC 8 · 2026 AMC 8