Difference between revisions of "2016 AMC 10A Problems/Problem 2"
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<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== | + | ==Solution 1== |
We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: | We can rewrite <math>10^{x}\cdot 100^{2x}=1000^{5}</math> as <math>10^{5x}=10^{15}</math>: | ||
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\end{split}</cmath> | \end{split}</cmath> | ||
Since the bases are equal, we can set the exponents equal, giving us <math>5x=15</math>. Solving the equation gives us <math>x = \boxed{\textbf{(C)}\;3}.</math> | Since the bases are equal, we can set the exponents equal, giving us <math>5x=15</math>. Solving the equation gives us <math>x = \boxed{\textbf{(C)}\;3}.</math> | ||
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+ | ==Solution 2== | ||
+ | |||
+ | We can rewrite this expression as <math>log(10^x \cdot 100^{2x})=log(1000^5)</math> , which can be simplified to <math>log(10^{x}\cdot10^{4x})=5log(1000)</math>, and that can be further simplified to <math>log(10^{5x})=5log(10^3)</math> . This leads to <math>5x=15</math>. Solving this linear equation yields <math>x = \boxed{\textbf{(C)}\;3}.</math> | ||
==See Also== | ==See Also== |
Revision as of 16:09, 17 March 2016
Contents
Problem
For what value of does
?
Solution 1
We can rewrite as
:
Since the bases are equal, we can set the exponents equal, giving us
. Solving the equation gives us
Solution 2
We can rewrite this expression as , which can be simplified to
, and that can be further simplified to
. This leads to
. Solving this linear equation yields
See Also
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.