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

Latest revision as of 14:43, 16 October 2025

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

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Video Solution

https://youtu.be/VIt6LnkV4_w?t=044

https://youtu.be/SZNMHoYzPDs

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2016 AMC 10A (Problems • Answer Key • Resources)
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 (Problems • Answer Key • Resources)
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

These problems are copyrighted © by the Mathematical Association of America, as part of the American Mathematics Competitions.

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 {{AMC12 box|year=2016|ab=A|num-b=1|num-a=3}}
 {{MAA Notice}}
+[[Category: Introductory Algebra Problems]]

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