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 $j+k.$

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

~IceMatrix