Difference between revisions of "2023 AIME I Problems/Problem 2"
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==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== |
Denote <math>x = \log_b n</math>. | Denote <math>x = \log_b n</math>. | ||
Hence, the system of equations given in the problem can be rewritten as | Hence, the system of equations given in the problem can be rewritten as | ||
<cmath> | <cmath> | ||
\begin{align*} | \begin{align*} | ||
| − | \sqrt{x} & = \frac{1}{2} x | + | \sqrt{x} & = \frac{1}{2} x , \\ |
bx & = 1 + x . | bx & = 1 + x . | ||
\end{align*} | \end{align*} | ||
</cmath> | </cmath> | ||
| − | + | Solving the system gives <math>x = 4</math> and <math>b = \frac{5}{4}</math>. | |
| − | |||
Therefore, | Therefore, | ||
| − | <cmath> | + | <cmath>n = b^x = \frac{625}{256}.</cmath> |
| − | |||
| − | n | ||
| − | |||
| − | |||
| − | </cmath> | ||
| − | |||
Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | Therefore, the answer is <math>625 + 256 = \boxed{881}</math>. | ||
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com) | ||
| − | ==Solution | + | ==Video Solution by TheBeautyofMath== |
| − | + | https://youtu.be/U96XHH23zhA | |
| + | |||
| + | ~IceMatrix | ||
==See also== | ==See also== | ||
{{AIME box|year=2023|num-b=1|num-a=3|n=I}} | {{AIME box|year=2023|num-b=1|num-a=3|n=I}} | ||
| + | {{MAA Notice}} | ||
| + | [[Category:Intermediate Algebra Problems]] | ||
Revision as of 13:54, 9 February 2024
Contents
Problem
Positive real numbers 
and 
satisfy the equations ![\[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\]](http://latex.artofproblemsolving.com/e/1/8/e188ed1d68418cfaed35d91cc7b0f333a84f6b1b.png)
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,
![\[n = b^x = \frac{625}{256}.\]](http://latex.artofproblemsolving.com/f/d/a/fda6b134d08ca7b814f36204c2f6a165af9ea24b.png)
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. 
