Difference between revisions of "2017 AMC 12A Problems/Problem 20"
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Since <math>\log_b a=x</math> is equivalent to <math>a=b^x</math>, each possible value <math>x</math> yields exactly <math>199</math> solutions <math>(b,a)</math>, as we can assign <math>a=b^x</math> to each <math>b=2,3,\dots,200</math>. In total, we have <math>3\cdot 199=\boxed{\textbf{(E) } 597}</math> solutions. | Since <math>\log_b a=x</math> is equivalent to <math>a=b^x</math>, each possible value <math>x</math> yields exactly <math>199</math> solutions <math>(b,a)</math>, as we can assign <math>a=b^x</math> to each <math>b=2,3,\dots,200</math>. In total, we have <math>3\cdot 199=\boxed{\textbf{(E) } 597}</math> solutions. | ||
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+ | ==Video Solution (HOW TO THINK CREATIVELY!!!)== | ||
+ | https://youtu.be/hBqRcXa_xqE | ||
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+ | ~Education, the Study of Everything | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2017|ab=A|num-b=19|num-a=21}} | {{AMC12 box|year=2017|ab=A|num-b=19|num-a=21}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 13:55, 10 June 2023
Problem
How many ordered pairs such that is a positive real number and is an integer between and , inclusive, satisfy the equation
Solution
By the properties of logarithms, we can rearrange the equation to read with . If , we may divide by it and get , which implies . Hence, we have possible values , namely
Since is equivalent to , each possible value yields exactly solutions , as we can assign to each . In total, we have solutions.
Video Solution (HOW TO THINK CREATIVELY!!!)
~Education, the Study of Everything
See Also
2017 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 19 |
Followed by Problem 21 |
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.