Art of Problem Solving
AoPS Online
Math texts, online classes, and more
for students in grades 5-12.
Visit AoPS Online ‚
Books for Grades 5-12 Online Courses
Beast Academy
Engaging math books and online learning
for students ages 6-13.
Visit Beast Academy ‚
Books for Ages 6-13 Beast Academy Online
AoPS Academy
Small live classes for advanced math
and language arts learners in grades 2-12.
Visit AoPS Academy ‚
Find a Physical Campus Visit the Virtual Campus

Difference between revisions of "2017 AMC 12A Problems/Problem 20"

(→‎Solution)
(→‎Solution)
Line 12: Line 12:
 
</cmath>
 
</cmath>
  
−
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,100</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.
  
 
==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 08:56, 23 January 2018

Problem

How many ordered pairs $(a,b)$ such that $a$ is a positive real number and $b$ is an integer between $2$ and $200$, inclusive, satisfy the equation $(\log_b a)^{2017}=\log_b(a^{2017})?$

$\textbf{(A)}\ 198\qquad\textbf{(B)}\ 199\qquad\textbf{(C)}\ 398\qquad\textbf{(D)}\ 399\qquad\textbf{(E)}\ 597$

Solution

By the properties of logarithms, we can rearrange the equation to read $x^{2017}=2017x$ with $x=\log_b a$. If $x\neq 0$, we may divide by it and get $x^{2016}=2017$, which implies $x=\pm \root{2016}\of{2017}$. Hence, we have $3$ possible values $x$, namely \[x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad x=-2017^{\frac1{2016}}.\]

Since $\log_b a=x$ is equivalent to $a=b^x$, each possible value $x$ yields exactly $199$ solutions $(b,a)$, as we can assign $a=b^x$ to each $b=2,3,\dots,200$. In total, we have $3\cdot 199=\boxed{\textbf{(E) } 597}$ solutions.

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. AMC logo.png

Retrieved from "https://artofproblemsolving.com/wiki/index.php?title=2017_AMC_12A_Problems/Problem_20&oldid=89935"