2016 AMC 10A Problems/Problem 2

Revision as of 15:09, 17 March 2016 by U2d2lrlrba ajia (talk | contribs) (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 1

We can rewrite $10^{x}\cdot 100^{2x}=1000^{5}$ as $10^{5x}=10^{15}$: \[\begin{split} 10^x\cdot100^{2x} & =10^x\cdot(10^2)^{2x} \\ 10^x\cdot10^{4x} & =(10^3)^5 \\ 10^{5x} & =10^{15} \end{split}\] Since the bases are equal, we can set the exponents equal, giving us $5x=15$. Solving the equation gives us $x = \boxed{\textbf{(C)}\;3}.$

Solution 2

We can rewrite this expression as $log(10^x \cdot 100^{2x})=log(1000^5)$ , which can be simplified to $log(10^{x}\cdot10^{4x})=5log(1000)$, and that can be further simplified to $log(10^{5x})=5log(10^3)$ . This leads to $5x=15$. Solving this linear equation yields $x = \boxed{\textbf{(C)}\;3}.$

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
2016 AMC 12A (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 12 Problems and Solutions

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