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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>
 
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>
  
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==Solution 1==
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==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
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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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==Solution 2 (extremely similar to above)==
 
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First, take the first equation and convert <math>\log_b\sqrt{n}</math> to <math>\log_b n^{\cfrac12}=\dfrac12\log_b n</math>. Square both sides to get <math>\log_b n=1/4 (\log_b n)^2</math>. Because a logarithm cannot be equal to <math>0</math>, <math>\log_b n=4</math>.
 
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By another logarithm rule, <math>\log_b(bn)=\log_b b+\log_b n=1+4=5</math>. Therefore, <math>4b=5</math>, and <math>b=\dfrac54</math>. Since <math>b^4=n</math>, we have <math>n=\dfrac{625}{256}</math>, and <math>a+b=\boxed{881}</math>.
 
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~wuwang2002 (feel free to remove if this is too similar to the above)
 
  
 
==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 02:52, 9 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*} Thus, $x = 4$ 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)

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

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