Difference between revisions of "2016 AMC 12A Problems/Problem 2"

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<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
 
<math>\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5</math>
  
==Solution==
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#REDIRECT [[2016 AMC 10A Problems/Problem 2]]
<cmath>10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}</cmath>
 
<cmath>10^x\cdot10^{4x}=(10^3)^5</cmath>
 
<cmath>10^{5x}=10^{15}</cmath>
 
<cmath>5x=15</cmath>
 
<cmath>x = \boxed{\textbf{(C) } \, 3}</cmath>
 
  
 
==See Also==
 
==See Also==
 
{{AMC12 box|year=2016|ab=A|num-b=1|num-a=3}}
 
{{AMC12 box|year=2016|ab=A|num-b=1|num-a=3}}
 
{{MAA Notice}}
 
{{MAA Notice}}

Revision as of 11:45, 4 February 2016

Problem

For what value of $x$ does $10^x\cdot100^{2x}=1000^5$?

$\textbf{(A)}\ 1\qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\textbf{(E)}\ 5$

  1. REDIRECT 2016 AMC 10A Problems/Problem 2

See Also

2016 AMC 12A (ProblemsAnswer KeyResources)
Preceded by
Problem 1
Followed by
Problem 3
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All AMC 12 Problems and Solutions

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