2023 AIME I Problems/Problem 2

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

Video Solution by TheBeautyofMath

https://youtu.be/U96XHH23zhA

~IceMatrix

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