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

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<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==
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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>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.
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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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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
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<math>b^{\sqrt[2016]{2017}}</math> must be a real number (It's a real number raised to a real number).
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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>.

Revision as of 14:54, 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 $2017log_b a$ from each side yields $(log_b a)^{2017}-2017log_b a=0$. We then proceed to factor out the term $log_b a$ which results in $(log_b a)[(log_b a)2016-2017]=0$. 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 $199$ solutions, as $b^{\sqrt[2016]{2017}}$ must be a real number (It's a real number raised to a real number).

Therefore, there are as many solutions as possible $b$ values, and as there is only one value of a for each $b$, $199 + 199 = 398$, therefore the answer is $\textbf{D}$.

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