Art of Problem Solving

2016 AMC 10A Problems/Problem 2: Difference between revisions

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==Solution==
==Solution==


We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>. Since the bases are equal, we can set the exponents equal: <math>5x=15</math>. Solving gives us: <cmath>x = \boxed{\textbf{(C)}\;3.}</cmath>
We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>. Since the bases are equal, we can set the exponents equal: <math>5x=15.</math> Solving gives us: <math>x = \boxed{\textbf{(C)}\;3.}</math>
 
 
==See Also==
{{AMC10 box|year=2016|ab=A|before=First Problem|num-a=2}}
{{MAA Notice}}

Revision as of 18:23, 3 February 2016

Problem

For what value of $x$ does $10^{x}\cdot 100^{2x}=1000^{5}$?

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

Solution

We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$. Since the bases are equal, we can set the exponents equal: $5x=15.$ Solving gives us: $x = \boxed{\textbf{(C)}\;3.}$


See Also

2016 AMC 10A (ProblemsAnswer KeyResources)
Preceded by
First Problem
Followed by
Problem 2
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

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