Difference between revisions of "2016 AMC 10A Problems/Problem 2"
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==Solution 2== | ==Solution 2== | ||
| − | We can rewrite this expression as <math>log(10^x \cdot 100^{2x})=log(1000^5)</math> , which can be simplified to <math>log(10^{x}\cdot10^{4x})= | + | We can rewrite this expression as <math>\log(10^x \cdot 100^{2x})=\log(1000^5)</math> , which can be simplified to <math>\log(10^{x}\cdot10^{4x})=5\log(1000)</math>, and that can be further simplified to <math>\log(10^{5x})=5\log(10^3)</math> . This leads to <math>5x=15</math>. Solving this linear equation yields <math>x = \boxed{\textbf{(C)}\;3}.</math> |
==See Also== | ==See Also== | ||
Revision as of 17:40, 17 March 2016
Contents
Problem
For what value of 
does 
?
Solution 1
We can rewrite as 
:
![\[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\]](http://latex.artofproblemsolving.com/7/9/d/79d64c5be65781d58e39c297545f5c2ceb9ba6f5.png)
Since the bases are equal, we can set the exponents equal, giving us 
. Solving the equation gives us 
Solution 2
We can rewrite this expression as 
, which can be simplified to 
, and that can be further simplified to 
. This leads to . Solving this linear equation yields
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. 
