Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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<math>2*\log_2 2x = 3*\log_2 3x</math> | <math>2*\log_2 2x = 3*\log_2 3x</math> | ||
− | Then <math>\log_2 (2x)^2 = \log_2 (3x)^3</math>. so <math>4x^2=27x^3</math> and we have <math>x=\frac{4}{27}</math> leading to <math>\boxed{\textbf{(D) 31 | + | Then <math>\log_2 (2x)^2 = \log_2 (3x)^3</math>. so <math>4x^2=27x^3</math> and we have <math>x=\frac{4}{27}</math> leading to <math>\boxed{\textbf{(D)}31}</math> (jeremylu) |
==Solution 2== | ==Solution 2== |
Revision as of 19:14, 18 August 2018
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 2*1 = 3
so now = putting in exponent form gets
=
so 4 = 27
dividing yields x = 4/27 and
4+27 =
- 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
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.