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2017 AMC 12A Problems/Problem 20

Revision as of 19:13, 8 February 2017 by Thedoge (talk | contribs) (→‎Problem)

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 $\textbf{D}$.

Note: this solution is 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

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