2017 AMC 12A Problems/Problem 20
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
.
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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