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

Revision as of 01:07, 27 October 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 $j+k.$ ok

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)

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

https://youtu.be/U96XHH23zhA

~IceMatrix

@@ Line 1: / Line 1: @@
 ==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> ok
 ==Solution==

2023 AIME I (Problems • Answer Key • Resources)
Preceded by Problem 1	Followed by Problem 3
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