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
We apply the Change of Base Formula, then rearrange: By the logarithmic identity it follows that from which the answer is
~jeremylu (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 2
By the logarithmic identity the original equation becomes By the logarithmic identity we multiply both sides by then apply the Change of Base Formula to the left side: Therefore, the answer is
~Pikachu13307 (Fundamental Logic)
~MRENTHUSIASM (Reconstruction)
Solution 4
We can convert both and into and respectively: Converting the bases of the right side, we get Dividing both sides by we get from which Expanding this equation gives Thus, we have from which the answer is
~lepetitmoulin (Solution)
~MRENTHUSIASM (Reformatting)
Solution 5
is the same as
Using Reciprocal law, we get
~OlutosinNGA
Solution 6
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 |
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 |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.