2017 AMC 12A Problems/Problem 20
How many ordered pairs such that is a positive real number and is an integer between and , inclusive, satisfy the equation
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 .