2016 AMC 10A Problems/Problem 2: Difference between revisions
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We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: | We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: | ||
<cmath>10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x} | <cmath>\begin{split} | ||
10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ | |||
10^x\cdot10^{4x} & =(10^3)^5 \\ | |||
Since the bases are equal, we can set the exponents equal | 10^{5x} & =10^{15} | ||
\end{split}</cmath> | |||
Since the bases are equal, we can set the exponents equal, giving us <math>5x=15</math>. Solving the equation gives us <math>x = \boxed{\textbf{(C)}\;3}.</math> | |||
==See Also== | ==See Also== | ||
Revision as of 16:43, 17 February 2016
Problem
For what value of 
does 
?
Solution
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 
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 | |
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