2010 AMC 8 Problems/Problem 24

Problem

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

$\textbf{(A)}\ 2^{24}<10^8<5^{12}\\ \textbf{(B)}\ 2^{24}<5^{12}<10^8 \\ \textbf{(C)}\ 5^{12}<2^{24}<10^8 \\ \textbf{(D)}\ 10^8<5^{12}<2^{24} \\ \textbf{(E)}\ 10^8<2^{24}<5^{12}$

Solution 2

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 $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.

Solution 3

First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations. $10^8$ is as fine as it is. We can rewrite $2^{24}$ as $(2^3)^8=8^8$ . Then we can rewrite $5^{12}$ as $(5^{\frac{3}{2}})^8=(\sqrt{125})^8)$ . We take the eighth root of all of these to get ${10, 8, \sqrt{125}}$ . Obviously, $8<10<\sqrt{125}$ , so the answer is $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ . Solution by Math

Solution 4

We know that $10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}$ . We also know that $5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}$ . If we remove the common factor of $5^{8}$ from both expressions, we are left with $2^{8}$ , which equals 256, and $5^{4}$ , which equals 625. So we know that $5^{12}$ is bigger than $10^{8}$ . This leaves us with only 2 options, A or E. Now we need to figure out which is bigger, $10^{8}$ or $2^{24}$ . To do this, we rewrite $2^{24}$ as $(2^{8})^{3}=(2^{3})^{8}=(8)^{8}$ , which is clearly less than $10^{8}$ . Therefore, $\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}$ is the correct answer.

By naman14

Video Solution by OmegaLearn

https://youtu.be/rQUwNC0gqdg?t=381

Video by MathTalks

https://youtu.be/mSCQzmfdX-g

Video Solution by WhyMath

https://youtu.be/EfCyJF1FEO ~someone

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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All AJHSME/AMC 8 Problems and Solutions

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