Difference between revisions of "2018 AMC 12A Problems/Problem 14"
Pikachu13307 (talk | contribs) (→Solution) |
m (missing references link (brides-maid)) |
||
Line 1: | Line 1: | ||
− | == Problem == | + | ==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 <math>\tfrac{1}{2}</math>, can be written as <math>\tfrac {p}{q}</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p + q</math>? | 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 <math>\tfrac{1}{2}</math>, can be written as <math>\tfrac {p}{q}</math> where <math>p</math> and <math>q</math> are relatively prime positive integers. What is <math>p + q</math>? | ||
Line 9: | Line 9: | ||
\textbf{(E) } 35 </math> | \textbf{(E) } 35 </math> | ||
− | == Solution == | + | ==Solution 1== |
Revision as of 15:10, 11 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 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
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.