Difference between revisions of "2023 AIME I Problems/Problem 2"

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~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
 
~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)
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==Video Solution 1 by TheBeautyofMath==
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https://youtu.be/U96XHH23zhA
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~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}}
 
{{MAA Notice}}

Revision as of 06:56, 13 February 2023

Problem

Positive real numbers $b \not= 1$ and $n$ satisfy the equations \[\sqrt{\log_b n} = \log_b \sqrt{n} \qquad \text{and} \qquad b \cdot \log_b n = \log_b (bn).\] The value of $n$ is $\frac{j}{k},$ where $j$ and $k$ are relatively prime positive integers. Find $j+k.$

Solution

Denote $x = \log_b n$. Hence, the system of equations given in the problem can be rewritten as \begin{align*} \sqrt{x} & = \frac{1}{2} x , \\ bx & = 1 + x . \end{align*} Solving the system gives $x = 4$ and $b = \frac{5}{4}$. Therefore, \[n = b^x = \frac{625}{256}.\] Therefore, the answer is $625 + 256 = \boxed{881}$.

~Steven Chen (Professor Chen Education Palace, www.professorchenedu.com)

Video Solution 1 by TheBeautyofMath

https://youtu.be/U96XHH23zhA

~IceMatrix

See also

2023 AIME I (ProblemsAnswer KeyResources)
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

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