Difference between revisions of "2016 AMC 10A Problems/Problem 2"
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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})=5\log(1000)</math>, and that can be further simplified to <math>\log(10^{5x})=5\log(10^3)</math> . This leads to <math>5x=15</math>. Solving this linear equation yields <math>x = \boxed{\textbf{(C)}\;3}.</math> | 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})=5\log(1000)</math>, and that can be further simplified to <math>\log(10^{5x})=5\log(10^3)</math> . This leads to <math>5x=15</math>. Solving this linear equation yields <math>x = \boxed{\textbf{(C)}\;3}.</math> | ||
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/-U6n2jBDqSA | ||
+ | |||
+ | ~Education, the Study of Everything | ||
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+ | |||
==Video Solution== | ==Video Solution== | ||
https://youtu.be/VIt6LnkV4_w?t=044 | https://youtu.be/VIt6LnkV4_w?t=044 | ||
− | ~ | + | https://youtu.be/SZNMHoYzPDs |
+ | |||
+ | ~savannahsolver | ||
==See Also== | ==See Also== |
Latest revision as of 09:58, 16 June 2023
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
Video Solution (HOW TO THINK CREATIVELY!!!)
https://youtu.be/-U6n2jBDqSA
~Education, the Study of Everything
Video Solution
https://youtu.be/VIt6LnkV4_w?t=044
~savannahsolver
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.