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"

m (→‎Problem)
(→‎Solution)
Line 7: Line 7:
 
==Solution==
 
==Solution==
  
−
By the properties of logarithms, we can rearrange the equation to read <math>2017 log_b a=(log_b a)^{2017}</math>. Then, subtracting <math>2017log_b a</math> from each side yields <math>(log_b a)^{2017}-2017log_b a=0</math>. We then proceed to factor out the term <math>log_b a</math> which results in <math>(log_b a)[(log_b a)2016-2017]=0</math>. Then, we set both factors equal to zero and solve.  
+
By the properties of logarithms, we can rearrange the equation to read <math>2017 \log_b a=(\log_b a)^{2017}</math>. Then, subtracting <math>2017\log_b a</math> from each side yields <math>(\log_b a)^{2017}-2017\log_b a=0</math>. We then proceed to factor out the term <math>\log_b a</math> which results in <math>(\log_b a)(2016\log_b a -2017)=0</math>. Then, we set both factors equal to zero and solve.  
  
−
<math>log_b a=0</math> has exactly <math>199</math> solutions with the restricted domain of <math>[2,200]</math> since this equation will always have a solution in the form of <math>(1, b)</math>, and there are <math>199</math> possible values of <math>b</math> since <math>200-2+1 = 199</math>.  
+
<math>\log_b a=0</math> has exactly <math>199</math> solutions with the restricted domain of <math>[2,200]</math> since this equation will always have a solution in the form of <math>(1, b)</math>, and there are <math>199</math> possible values of <math>b</math> since <math>200-2+1 = 199</math>.  
  
−
We proceed to solve the other factor, <math>(log_b a)2016-2017</math>. We add <math>2017</math> to both sides, and take the <math>2016th</math> root, this gives us <math>log_b a=\sqrt[2016]{2017}</math> <math>\sqrt[2016]{2017}</math> is a real number, and therefore <math>a=b^{\sqrt[2016]{2017}}</math> Again, there are <math>199</math> solutions, as  
+
We proceed to solve the other factor, <math>(\log_b a)2016-2017</math>. We add <math>2017</math> to both sides, and take the <math>2016th</math> root, this gives us <math>\log_b a=\sqrt[2016]{2017}</math> <math>\sqrt[2016]{2017}</math> is a real number, and therefore <math>a=b^{\sqrt[2016]{2017}}</math> Again, there are <math>199 \cdot 2 = 398</math> solutions, as <math>b^{\sqrt[2016]{2017}}</math> must be a real number (It's a real number raised to a real number), and we also have to multiply by two because when we take the <math>2016th</math> root, we must also consider the negative root which is valid because the taking the reciprocal of <math>a</math> negates <math>log_ba</math>.
−
<math>b^{\sqrt[2016]{2017}}</math> must be a real number (It's a real number raised to a real number).  
 
  
−
Therefore, there are as many solutions as possible <math>b</math> values, and as there is only one value of a for each <math>b</math>, <math>199 + 199 = 398</math>, therefore the answer is <math>\textbf{D}</math>.
+
Therefore, the answer is <math>199 + 398 = \boxed{\textbf{E } 597}</math>.
  
−
Note: this solution is incorrect because when we take the <math>2016th</math> root, we must also consider the negative root which is valid because the taking the reciprocal of <math>a</math> negates <math>log_ba</math>. Therefore the answer is <math>199 \cdot 3</math> or <math>\textbf{E}</math>.
+
Note: the former solution was incorrect because when we take the <math>2016th</math> root, we must also consider the negative root which is valid because the taking the reciprocal of <math>a</math> negates <math>log_ba</math>. Therefore the answer is <math>199 \cdot 3</math> or <math>\textbf{E}</math>.
  
 
==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 19:17, 8 February 2017

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 $2017 \log_b a=(\log_b a)^{2017}$. Then, subtracting $2017\log_b a$ from each side yields $(\log_b a)^{2017}-2017\log_b a=0$. We then proceed to factor out the term $\log_b a$ which results in $(\log_b a)(2016\log_b a -2017)=0$. Then, we set both factors equal to zero and solve.

$\log_b a=0$ has exactly $199$ solutions with the restricted domain of $[2,200]$ since this equation will always have a solution in the form of $(1, b)$, and there are $199$ possible values of $b$ since $200-2+1 = 199$.

We proceed to solve the other factor, $(\log_b a)2016-2017$. We add $2017$ to both sides, and take the $2016th$ root, this gives us $\log_b a=\sqrt[2016]{2017}$ $\sqrt[2016]{2017}$ is a real number, and therefore $a=b^{\sqrt[2016]{2017}}$ Again, there are $199 \cdot 2 = 398$ solutions, as $b^{\sqrt[2016]{2017}}$ must be a real number (It's a real number raised to a real number), and we also have to multiply by two because when we take the $2016th$ root, we must also consider the negative root which is valid because the taking the reciprocal of $a$ negates $log_ba$.

Therefore, the answer is $199 + 398 = \boxed{\textbf{E } 597}$.

Note: the former solution was incorrect because when we take the $2016th$ root, we must also consider the negative root which is valid because the taking the reciprocal of $a$ negates $log_ba$. Therefore the answer is $199 \cdot 3$ or $\textbf{E}$.

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=83146"