2018 AMC 12A Problems/Problem 14
Problem
The solutions to the equation , where
is a positive real number other than
or
, can be written as
where
and
are relatively prime positive integers. What is
?
Solution 1
Base switch to log 2 and you have .
Then . so
and we have
leading to
(jeremylu)
Solution 2
If you multiply both sides by
then it should come out to *
=
*
that then becomes *
=
*
which simplifies to
so now =
putting in exponent form gets
=
so =
dividing yields and
- Pikachu13307
Solution 3
We can convert both and
into
and
, respectively, giving:
Converting the bases of the right side, we get
Dividing both sides by , we get
Which simplifies to
Log expansion allows us to see that
, which then simplifies to
Thus,
And
-lepetitmoulin
Solution 4
is the same as
Using Reciprocal law, we get
~OlutosinNGA
Solution 5
. We know that
. Thus
.
and
are indeed relatively prime thus our final answer is
-vsamc
See Also
2018 AMC 12A (Problems • Answer Key • Resources) | |
Preceded by Problem 13 |
Followed by Problem 15 |
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All AMC 12 Problems and Solutions |
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