Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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that then becomes <math>\log_2 (4)</math> * <math>\log_{3x} (3x)</math> = <math>\log_2 (8)</math> * <math>\log_{2x} (3x)</math> | that then becomes <math>\log_2 (4)</math> * <math>\log_{3x} (3x)</math> = <math>\log_2 (8)</math> * <math>\log_{2x} (3x)</math> | ||
− | which simplifies to 2*1 = 3 | + | which simplifies to <math>2*1 = 3\log_{2x} (3x)</math> |
so now <math>\frac{2}{3}</math> = <math>\log_{2x} (3x)</math> putting in exponent form gets | so now <math>\frac{2}{3}</math> = <math>\log_{2x} (3x)</math> putting in exponent form gets | ||
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<math>(2x)^2</math> = <math>(3x)^3</math> | <math>(2x)^2</math> = <math>(3x)^3</math> | ||
− | so | + | so <math>4x^2</math> = <math>27x^3</math> |
− | dividing yields x = 4/27 and | + | dividing yields <math>x = 4/27</math> and |
− | 4+27 = <math>\boxed{\textbf{(D)}31}</math> | + | <math>4+27 =</math> <math>\boxed{\textbf{(D)}31}</math> |
- Pikachu13307 | - Pikachu13307 | ||
Revision as of 03:55, 25 January 2019
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
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 |
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