Difference between revisions of "2021 Fall AMC 10B Problems/Problem 5"
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==Solution 1== | ==Solution 1== | ||
| − | We have <cmath>n=8^{2022}= \left(8^\frac{2}{3}\right)^{2022}=4^{3033}</cmath> | + | We have <cmath>n=8^{2022}= \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.</cmath> |
| − | Therefore, <cmath>\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}</cmath> | + | Therefore, <cmath>\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}.</cmath> |
~kingofpineapplz | ~kingofpineapplz | ||
Revision as of 20:54, 22 November 2021
Problem 5
Let 
. Which of the following is equal to 
Solution 1
We have ![\[n=8^{2022}= \left(8^\frac{2}{3}\right)^{2022}=4^{3033}.\]](http://latex.artofproblemsolving.com/6/b/f/6bff93b33bc90f12c694a4edf3d2587b9aab71c8.png)
Therefore, ![\[\frac{n}4=\boxed{(\textbf{E})\:4^{3032}}.\]](http://latex.artofproblemsolving.com/3/3/b/33badf2380db5fd9beb44c3dcc25766267038e1e.png)
~kingofpineapplz
