Difference between revisions of "2018 AMC 12A Problems/Problem 14"
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4+27 = <math>\boxed{ (D) 31}</math> | 4+27 = <math>\boxed{ (D) 31}</math> | ||
- Pikachu13307 | - Pikachu13307 | ||
+ | |||
+ | ==Solution 3== | ||
+ | |||
+ | We can convert both <math>4</math> and <math>8</math> into <math>2^2</math> and <math>2^3</math>, respectively, giving: | ||
+ | |||
+ | <math>2\log_{3x} (2) = 3\log_{2x} (2)</math> | ||
+ | |||
+ | Converting the bases of the right side, we get <math>\log_{2x} 2 = \frac{\ln 2}{\ln (2x)}</math> | ||
+ | |||
+ | <math>\frac{2}{3}*\log_{3x} (2) = \frac{\ln 2}{\ln (2x)}</math> | ||
+ | |||
+ | <math>2^\frac{2}{3} = (3x)^\frac{\ln 2}{\ln (2x)}</math> | ||
+ | |||
+ | <math>\frac{2}{3} * \ln 2 = \frac{\ln 2}{\ln (2x)} * \ln (3x)</math> | ||
+ | |||
+ | Dividing both sides by <math>\ln 2</math>, we get | ||
+ | |||
+ | <math>\frac{2}{3} = \frac{\ln (3x)}{\ln (2x)}</math> | ||
+ | |||
+ | Which simplifies to | ||
+ | |||
+ | <math>2\ln (2x) = 3\ln (3x)</math> | ||
+ | |||
+ | Log expansion allows us to see that | ||
+ | |||
+ | <math>2\ln 2 + 2\ln (x) = 3\ln 3 + 3\ln (x)</math>, which then simplifies to | ||
+ | |||
+ | <math>\ln (x) = 2\ln 2 - 3\ln 3</math> | ||
+ | |||
+ | Eventually, we find that | ||
+ | |||
+ | <math>x = e^{2\ln 2 - 3\ln 3} = \frac{e^{2\ln 2}}{e^{3\ln 3}}</math> | ||
+ | |||
+ | And | ||
+ | |||
+ | <math>x = \frac{2^2}{3^3} = \frac{4}{27} = \boxed{ (D) 31}</math> | ||
+ | |||
+ | -lepetitmoulin | ||
==See Also== | ==See Also== | ||
{{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}} | {{AMC12 box|year=2018|ab=A|num-b=13|num-a=15}} | ||
{{MAA Notice}} | {{MAA Notice}} |
Revision as of 15:42, 23 March 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
Eventually, we find that
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.