Difference between revisions of "2010 AMC 8 Problems/Problem 24"

Latest revision as of 07:12, 9 November 2024

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 1

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 2

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 3

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 $10^{8}$ . Now we need to figure out which is bigger, $10^{8}$ or $2^{24}$ . To do this, we rewrite $2^{24}$ as $(2^{3})^{8}=(8)^{8}$ , 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

@@ Line 1: / Line 1: @@
-A
+==Problem==
+What is the correct ordering of the three numbers, <math>10^8</math>, <math>5^{12}</math>, and <math>2^{24}</math>?
+<math> \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} </math>
+== Solution 1==
+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 <math>10^2=100</math>, <math>5^3=125</math>, and <math>2^6=64</math>. Since <math>64<100<125</math>, it follows that <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> is the correct answer.
+== Solution 2==
+First, let us make all exponents equal to 8. Then, it will be easy to order the numbers without doing any computations.
+<math>10^8</math> is as fine as it is.
+We can rewrite <math>2^{24}</math> as <math>(2^3)^8=8^8</math>.
+Then we can rewrite <math>5^{12}</math> as <math>(5^{\frac{3}{2}})^8=(\sqrt{125})^8)</math>.
+We take the eighth root of all of these to get <math>{10, 8, \sqrt{125}}</math>.
+Obviously, <math>8<10<\sqrt{125}</math>, so the answer is <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math>.
+Solution by Math
+==Solution 3==
+We know that <math>10^{8}=({5}\cdot{2})^{8}=5^{8}\cdot2^{8}</math>. We also know that <math>5^{12}=5^{(8+4)}=5^{8}\cdot5^{4}</math>. If we remove the common factor of <math>5^{8}</math> from both expressions, we are left with <math>2^{8}</math>, which equals 256, and <math>5^{4}</math>, which equals 625. So we know that <math>5^{12}</math> is bigger than <math>10^{8}</math>. Now we need to figure out which is bigger, <math>10^{8}</math> or <math>2^{24}</math>. To do this, we rewrite <math>2^{24}</math> as <math>(2^{3})^{8}=(8)^{8}</math>, which is clearly less than <math>10^{8}</math>. Therefore, <math>\boxed{\textbf{(A)}\ 2^{24}<10^8<5^{12}}</math> 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
+==See Also==
+{{AMC8 box|year=2010|num-b=23|num-a=25}}
+{{MAA Notice}}

2010 AMC 8 (Problems • Answer Key • Resources)
Preceded by Problem 23	Followed by Problem 25
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