Difference between revisions of "2023 AIME I Problems/Problem 2"
Mathkiddie (talk | contribs) (→Solution 2) |
Megahertz13 (talk | contribs) (→Problem) |
||
Line 1: | Line 1: | ||
==Problem== | ==Problem== | ||
Positive real numbers <math>b \not= 1</math> and <math>n</math> satisfy the equations <cmath>\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).</cmath> The value of <math>n</math> is <math>\frac{j}{k},</math> where <math>j</math> and <math>k</math> are relatively prime positive integers. Find <math>j+k.</math> | Positive real numbers <math>b \not= 1</math> and <math>n</math> satisfy the equations <cmath>\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).</cmath> The value of <math>n</math> is <math>\frac{j}{k},</math> where <math>j</math> and <math>k</math> are relatively prime positive integers. Find <math>j+k.</math> | ||
+ | ==Video Solution & More by MegaMath== | ||
+ | https://www.youtube.com/watch?v=jxY7BBe-4gU | ||
==Solution== | ==Solution== |
Revision as of 12:54, 9 February 2024
Contents
[hide]Problem
Positive real numbers and satisfy the equations The value of is where and are relatively prime positive integers. Find
Video Solution & More by MegaMath
https://www.youtube.com/watch?v=jxY7BBe-4gU
Solution
Denote . Hence, the system of equations given in the problem can be rewritten as Solving the system gives and . Therefore, Therefore, the answer is .
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
Video Solution by TheBeautyofMath
~IceMatrix
See also
2023 AIME I (Problems • Answer Key • Resources) | ||
Preceded by Problem 1 |
Followed by Problem 3 | |
1 • 2 • 3 • 4 • 5 • 6 • 7 • 8 • 9 • 10 • 11 • 12 • 13 • 14 • 15 | ||
All AIME Problems and Solutions |
The problems on this page are copyrighted by the Mathematical Association of America's American Mathematics Competitions.