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

Revision as of 15:45, 14 January 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 $j+k.$

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)

Solution 2

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 19: / Line 19: @@
 ==Solution 2==
-Denote b=n^x. Hence, the system of equations given in the problem can be rewritten as
+Denote <math>b=n^x</math>. Hence, the system of equations given in the problem can be rewritten as
-sqrt(x)=x/2,  b*x=1+x
+<math>sqrt(x)=x/2</math>,  <math>b*x=1+x</math>
-Thus, x=x^2/4, x=4. So, n=b^4
+Thus, <math>x=x^2/4</math>, <math>x=4</math>. So, <math>n=b^4</math>
-Then, 4b=1+4. So, b=5/4. Then, n=625/256
+Then, <math>4b=1+4</math>. So, <math>b=5/4</math>. Then, <math>n=625/256</math>
 Ans=881

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