Difference between revisions of "2016 AMC 10A Problems/Problem 2"

(Created page with "For what value of <math>x</math> does <math>10^{x}\cdot 100^{2x}=1000^{5}</math>? <math>\textbf{(A)}\ 1 \qquad\textbf{(B)}\ 2\qquad\textbf{(C)}\ 3\qquad\textbf{(D)}\ 4\qquad\...")
 
Line 2: Line 2:
  
 
<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==
 +
 +
We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>. From here, we can easily solve for <math>x</math>, and <cmath>x = \boxed{\textbf{(C)}\;3.}</cmath>

Revision as of 19:10, 3 February 2016

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 $10^{5x}=10^{15}$. From here, we can easily solve for $x$, and \[x = \boxed{\textbf{(C)}\;3.}\]