Difference between revisions of "2023 AIME I Problems/Problem 2"
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+ | 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> | ||
==Solution== | ==Solution== |
Revision as of 18:19, 27 October 2023
Problem
Positive real numbers and satisfy the equations The value of is where and are relatively prime positive integers. Find
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)
Solution2
Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as sqrt(x)=x/2, b*x=1+x Thus, x=x^2/4, x=4. So, n=b^4 Then, 4b=1+4. So, b=5/4. Then, n=625/256 Ans=881
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.