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

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==Solution==
 
==Solution==
  
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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)((\log_b a)^{2016} -2017)=0</math>. Then, we set both factors equal to zero and solve.  
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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
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<cmath>
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x=0,\qquad x=2017^{\frac1{2016}},\, \text{and}\quad  x=-2017^{\frac1{2016}}.
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</cmath>
  
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<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>.
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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,100</math>. In total, we have <math>3\cdot 199=\boxed{\textbf{(E) } 597}</math> solutions.
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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>.
 
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Therefore, the answer is <math>199 + 398 = \boxed{\textbf{(E) } 597}</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 16:01, 13 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 $x^{2017}=2017x$ 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 $b=2,3,\dots,100$. 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

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