Difference between revisions of "2016 AMC 10A Problems/Problem 2"
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==See Also== | ==See Also== | ||
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| + | {{AMC12 box|year=2016|ab=A|num-b=1|num-a=3}} | ||
{{MAA Notice}} | {{MAA Notice}} | ||
Revision as of 11:46, 4 February 2016
Problem
For what value of 
does 
?
Solution
We can rewrite as 
:
![\[10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}\]](http://latex.artofproblemsolving.com/d/5/1/d51f842b39abedd9cf4fb239e9fc69f14a9e0ad2.png)
![\[10^x\cdot10^{4x}=(10^3)^5\]](http://latex.artofproblemsolving.com/7/b/5/7b5d31779821c75b6d4360a93ff0f2994cb278b2.png)
![\[10^{5x}=10^{15}\]](http://latex.artofproblemsolving.com/4/8/e/48e7dee4371b7871eee2a1daf110fc82702bd64b.png)
Since the bases are equal, we can set the exponents equal: 
. Solving gives us: 
See Also
| 2016 AMC 10A (Problems • Answer Key • Resources) | ||
| Preceded by Problem 1 |
Followed by Problem 3 | |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | ||
| All AMC 10 Problems and Solutions | ||
| 2016 AMC 12A (Problems • Answer Key • Resources) | |
| Preceded by Problem 1 |
Followed by Problem 3 |
| 1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 • 16 • 17 • 18 • 19 • 20 • 21 • 22 • 23 • 24 • 25 | |
| All AMC 12 Problems and Solutions | |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. 
