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

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==Problem==
 
==Problem==
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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> ok
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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>
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==Video Solution & More by MegaMath==
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https://www.youtube.com/watch?v=jxY7BBe-4gU
  
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==Solution==
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==Solution 1==
 
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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==Solution2==
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==Solution 2==
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Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as
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We can use the property that <math>\log(xy) = \log(x) + \log(y)</math> on the first equation. We get <math>b \log_b(n) = 1 + \log_b(n)</math>. Then, subtracting <math>\log_b(n)</math> from both sides, we get <math>(b-1) \log_b(n) = 1</math>, therefore <math>\log_b(n) = \frac{1}{b-1}</math>. Substituting that into our first equation, we get <math>\frac{1}{2b-2} = \sqrt{\frac{1}{b-1}}</math>. Squaring, reciprocating, and simplifying both sides, we get the quadratic <math>4b^2 - 9b + 5 = 0</math>. Solving for <math>b</math>, we get <math>\frac{5}{4}</math> and <math>1</math>. Since the problem said that <math>b \neq 1</math>, <math>b = \frac{5}{4}</math>. To solve for <math>n</math>, we can use the property that <math>\log_b(n) = \frac{1}{b-1}</math>. <math>\log_\frac{5}{4}(n) = 4</math>, so <math>n = \frac{5^4}{4^4} = \frac{625}{256}</math>. Adding these together, we get <math>\boxed{881}</math>
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sqrt(x)=x/2, b*x=1+x
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Thus, x=x^2/4, x=4. So, n=b^4
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~idk12345678
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Then, 4b=1+4. So, b=5/4. Then, n=625/256
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Ans=881
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==Solution 3 (quick)==
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We can let <math>n=b^{4x^2}</math>. Then, in the first equation, the LHS becomes <math>2x</math> and the RHS becomes <math>2x^2</math>. Therefore, <math>x</math> must be <math>1</math> (<math>x</math> can't be <math>0</math>). So now we know <math>n=b^4</math>. So we can plug this into the second equation to get <math>n=b^4</math>. This gives <math>b\cdot4=5</math>, so <math>b= \frac{5}{4}</math> and <math>n= b^4=\frac{625}{256}</math>. Adding the numerator and denominator gives <math>\boxed{881}</math>.
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Honestly this problem is kinda misplaced.
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~yrock
  
 
==Video Solution by TheBeautyofMath==
 
==Video Solution by TheBeautyofMath==

Latest revision as of 19:46, 23 May 2024

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 $j+k.$

Video Solution & More by MegaMath

https://www.youtube.com/watch?v=jxY7BBe-4gU

Solution 1

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)

Solution 2

We can use the property that $\log(xy) = \log(x) + \log(y)$ on the first equation. We get $b \log_b(n) = 1 + \log_b(n)$. Then, subtracting $\log_b(n)$ from both sides, we get $(b-1) \log_b(n) = 1$, therefore $\log_b(n) = \frac{1}{b-1}$. Substituting that into our first equation, we get $\frac{1}{2b-2} = \sqrt{\frac{1}{b-1}}$. Squaring, reciprocating, and simplifying both sides, we get the quadratic $4b^2 - 9b + 5 = 0$. Solving for $b$, we get $\frac{5}{4}$ and $1$. Since the problem said that $b \neq 1$, $b = \frac{5}{4}$. To solve for $n$, we can use the property that $\log_b(n) = \frac{1}{b-1}$. $\log_\frac{5}{4}(n) = 4$, so $n = \frac{5^4}{4^4} = \frac{625}{256}$. Adding these together, we get $\boxed{881}$

~idk12345678

Solution 3 (quick)

We can let $n=b^{4x^2}$. Then, in the first equation, the LHS becomes $2x$ and the RHS becomes $2x^2$. Therefore, $x$ must be $1$ ($x$ can't be $0$). So now we know $n=b^4$. So we can plug this into the second equation to get $n=b^4$. This gives $b\cdot4=5$, so $b= \frac{5}{4}$ and $n= b^4=\frac{625}{256}$. Adding the numerator and denominator gives $\boxed{881}$.

Honestly this problem is kinda misplaced.

~yrock

Video Solution by TheBeautyofMath

https://youtu.be/U96XHH23zhA

~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

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