Difference between revisions of "2017 AMC 12A Problems/Problem 20"
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==Problem== | ==Problem== | ||
− | How many ordered pairs <math>(a,b)</math> such that <math>a</math> is a positive real number and <math>b</math> is an integer between <math>2</math> and <math>200</math>, inclusive, satisfy the equation <math>(log_b a)^{2017}=log_b(a^{2017})?</math> | + | How many ordered pairs <math>(a,b)</math> such that <math>a</math> is a positive real number and <math>b</math> is an integer between <math>2</math> and <math>200</math>, inclusive, satisfy the equation <math>(\log_b a)^{2017}=\log_b(a^{2017})?</math> |
<math>\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597</math> | <math>\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597</math> | ||
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==Solution== | ==Solution== | ||
− | By the properties of logarithms, we can rearrange the equation to read <math> | + | By the properties of logarithms, we can rearrange the equation to read <math>x^{2017}=2017x</math> with <math>x=\log_b a</math>. If <math>x\neq 0</math>, we may divide by it and get <math>x^{2016}=2017</math>, which implies <math>x=\pm \root{2016}\of{2017}</math>. Hence, we have <math>3</math> possible values <math>x</math>, namely |
+ | <cmath> | ||
+ | x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad x=-2017^{\frac1{2016}}. | ||
+ | </cmath> | ||
− | <math>log_b a= | + | 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 BETTER!!!)== | |
+ | https://youtu.be/hBqRcXa_xqE | ||
− | + | ~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}} |
Latest revision as of 22:34, 16 September 2024
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 BETTER!!!)
~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 |
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