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

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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>. | 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>. | ||

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+ | 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>. |

## Revision as of 18:13, 8 February 2017

## Problem

How many ordered pairs such that is a positive real number and is an integer between and , inclusive, satisfy the equation

## Solution

By the properties of logarithms, we can rearrange the equation to read . Then, subtracting from each side yields . We then proceed to factor out the term which results in . Then, we set both factors equal to zero and solve.

has exactly solutions with the restricted domain of since this equation will always have a solution in the form of , and there are possible values of since .

We proceed to solve the other factor, . We add to both sides, and take the root, this gives us is a real number, and therefore Again, there are solutions, as must be a real number (It's a real number raised to a real number).

Therefore, there are as many solutions as possible values, and as there is only one value of a for each , , therefore the answer is .

Note: this solution is incorrect because when we take the root, we must also consider the negative root which is valid because the taking the reciprocal of negates . Therefore the answer is or .