2016 AMC 10A Problems/Problem 2

Revision as of 12:46, 4 February 2016 by Swe1 (talk | contribs) (Added some elements from old 12A solution)

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 $10^{5x}=10^{15}$: \[10^x\cdot100^{2x}=10^x\cdot(10^2)^{2x}\] \[10^x\cdot10^{4x}=(10^3)^5\] \[10^{5x}=10^{15}\] 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
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

The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions. AMC logo.png