Difference between revisions of "2018 AMC 12A Problems/Problem 14"
(Created page with "== Problem == The solutions to the equation <math>\log_{3x} 4 = \log_{2x} 8</math>, where <math>x</math> is a positive real number other than <math>\tfrac{1}{3}</math> or <ma...") |
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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{ (D) 31}</math> (jeremylu) | 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{ (D) 31}</math> (jeremylu) | ||
+ | ==Solution 2== | ||
+ | If you multiply both sides by <math>\log_2 (3x)</math> | ||
+ | |||
+ | then it should come out to <math>\log_2 (3x)</math> * <math>\log_{3x} (4)</math> = <math>\log_2 {3x}</math> * <math>\log_{2x} (8)</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<math>\log_{2x} (3x)</math> | ||
+ | |||
+ | so now <math>\frac{2}{3}</math> = <math>\log_{2x} (3x)</math> putting in exponent form gets | ||
+ | |||
+ | <math>(2x)^2</math> = <math>(3x)^3</math> | ||
+ | |||
+ | so 4<math>x^2</math> = 27<math>x^3</math> | ||
+ | |||
+ | dividing yields x = 4/27 and | ||
+ | |||
+ | 4+27 = <math>\boxed{ (D) 31}</math> | ||
+ | - Pikachu13307 | ||
==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 11:56, 9 February 2018
Contents
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
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
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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